6.4: Quadratic Functions and Their Graphs (2024)

Example \(\PageIndex{1}\):

Graph: \(f(x)=-x^{2}-2 x+3\).

Solution

Step 1: Determine the \(y\)-intercept. To do this, set \(x\)=0 and find \(f(0)\).

\(\begin{aligned} f(x) &=-x^{2}-2 x+3 \\ f(0) &=-(\color{OliveGreen}{0}\color{black}{)}^{2}-2(\color{OliveGreen}{0}\color{black}{)}+3 \\ &=3 \end{aligned}\)

The \(y\)-intercept is \((0,3)\).

Step 2: Determine the \(x\)-intercepts if any. To do this, set \(f(x)=0\) and solve for \(x\).

\(\begin{aligned} f(x)&=-x^{2}-2 x+3 \quad\:\color{Cerulean}{ Set\: f(x)=0. }\\ 0&=-x^{2}-2 x+3 \quad\:\color{Cerulean} { Multiply\: both\: sides\: by\: -1.} \\ 0&=x^{2}+2 x-3 \quad\:\:\:\:\color{Cerulean} { Factor. } \\ 0&=(x+3)(x-1) \:\:\:\color{Cerulean} { Set\: each\: factor\: equal\: to\: zero. }\end{aligned}\)

\(\begin{array}{rl}{x+3=0} & {\text { or } x-1=0} \\ {x=-3} & \quad\quad\quad{x=1}\end{array}\)

Here where \(f (x) = 0\), we obtain two solutions. Hence, there are two \(x\)-intercepts, \((−3, 0)\) and \((1, 0)\).

Step 3: Determine the vertex. One way to do this is to first use \(x = −\frac{b}{2a}\) to find the \(x\)-value of the vertex and then substitute this value in the function to find the corresponding \(y\)-value. In this example, \(a = −1\) and \(b = −2\).

\(\begin{aligned} x &=\frac{-b}{2 a} \\ &=\frac{-(\color{OliveGreen}{-2}\color{black}{)}}{2(\color{OliveGreen}{-1}\color{black}{)}} \\ &=\frac{2}{-2} \\ &=-1 \end{aligned}\)

Substitute \(−1\) into the original function to find the corresponding \(y\)-value.

\(\begin{aligned} f(x) &=-x^{2}-2 x+3 \\ f(\color{OliveGreen}{-1}\color{black}{)} &=-(\color{OliveGreen}{-1}\color{black}{)}^{2}-2(\color{OliveGreen}{-1}\color{black}{)}+3 \\ &=-1+2+3 \\ &=4 \end{aligned}\)

The vertex is \((-1,4)\).

Step 4: Determine extra points so that we have at least five points to plot. Ensure a good sampling on either side of the line of symmetry. In this example, one other point will suffice. Choose \(x = −2\) and find the corresponding \(y\)-value.

Our fifth point is \((−2, 3)\).

Step 5: Plot the points and sketch the graph. To recap, the points that we have found are

\(y\)-intercept: \((0,3)\)

\(x\)-intercepts: \((-3,0)\) and \((1,0)\)

Vertex: \((-1,4)\)

Extra point: \((-2,3)\)

Answer:

Example \(\PageIndex{2}\):

Graph: \(f(x)=2 x^{2}+4 x+5\).

Solution

Because the leading coefficient \(2\) is positive, we note that the parabola opens upward. Here \(c = 5\) and the \(y\)-intercept is \((0, 5)\). To find the \(x\)-intercepts, set \(f (x) = 0\).

\(\begin{aligned}f(x)=2 x^{2}+4 x+5 \\ 0=2 x^{2}+4 x+5\end{aligned}\)

In this case, \(a = 2, b = 4\), and \(c = 5\). Use the discriminant to determine the number and type of solutions.

\(\begin{aligned} b^{2}-4 a c &=(4)^{2}-4(2)(5) \\ &=16-40 \\ &=-24 \end{aligned}\)

Since the discriminant is negative, we conclude that there are no real solutions. Because there are no real solutions, there are no \(x\)-intercepts. Next, we determine the \(x\)-value of the vertex.

\(\begin{aligned} x &=\frac{-b}{2 a} \\ &=\frac{-(\color{OliveGreen}{4}\color{black}{)}}{2(\color{OliveGreen}{2}\color{black}{)}} \\ &=\frac{-4}{4} \\ &=-1 \end{aligned}\)

Given that the \(x\)-value of the vertex is \(−1\), substitute \(−1\) into the original equation to find the corresponding \(y\)-value.

\(\begin{aligned} f(x) &=2 x^{2}+4 x+5 \\ f(\color{OliveGreen}{-1}\color{black}{)} &=2(\color{OliveGreen}{-1}\color{black}{)}^{2}+4(\color{OliveGreen}{-1}\color{black}{)}+5 \\ &=2-4+5 \\ &=3 \end{aligned}\)

The vertex is \((−1, 3)\). So far, we have only two points. To determine three more, choose some \(x\)-values on either side of the line of symmetry, \(x = −1\). Here we choose \(x\)-values \(−3, −2\), and \(1\).

To summarize, we have

\(y\)-intercept: \((0,5)\)

\(x\)-intercepts: None

Vertex: \((-1,3)\)

Extra points: \((-3,11), (-2,5), (1,11)\)

Plot the points and sketch the graph.

Answer:

Example \(\PageIndex{3}\):

Graph: \(f(x)=x^{2}-2 x-1\).

Solution

Since \(a = 1\), the parabola opens upward. Furthermore, \(c = −1\), so the \(y\)-intercept is \((0, −1)\). To find the \(x\)-intercepts, set \(f (x) = 0\).

\(\begin{aligned} f(x) &=x^{2}-2 x-1 \\ 0 &=x^{2}-2 x-1 \end{aligned}\)

In this case, solve using the quadratic formula with \(a = 1, b = −2\), and \(c = −1\).

\(\begin{aligned} x &=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a} \\ &=\frac{-(\color{OliveGreen}{-2}\color{black}{)} \pm \sqrt{(\color{OliveGreen}{-2}\color{black}{)}^{2}-4(\color{OliveGreen}{1}\color{black}{)}(\color{OliveGreen}{-1}\color{black}{)}}}{2(\color{OliveGreen}{1}\color{black}{)}} \\ &=\frac{2 \pm \sqrt{8}}{2} \\ &=\frac{2 \pm 2 \sqrt{2}}{2} \\ &= \frac{2(1\pm\sqrt{2})}{2} \\ &=1 \pm \sqrt{2} \end{aligned}\)

Here we obtain two real solutions for \(x\), and thus there are two \(x\)-intercepts:

\(\begin{array}{c}{(1-\sqrt{2}, 0) \text { and }(1+\sqrt{2}, 0)} \quad\color{Cerulean}{Exact\:values} \\ \quad\quad\quad\quad{(-0.41,0) \quad\quad(2.41,0)}\quad\quad\color{Cerulean}{Approximate\:values}\end{array}\)

Approximating the \(x\)-intercepts using a calculator will help us plot the points. However, we will present the exact \(x\)-intercepts on the graph. Next, find the vertex.

\(\begin{aligned} x &=\frac{-b}{2 a} \\ &=\frac{-(\color{OliveGreen}{-2}\color{black}{)}}{2(\color{OliveGreen}{1}\color{black}{)}} \\ &=\frac{2}{2} \\ &=1 \end{aligned}\)

Given that the \(x\)-value of the vertex is \(1\), substitute into the original equation to find the corresponding \(y\)-value.

\(\begin{aligned} y &=x^{2}-2 x-1 \\ &=(\color{OliveGreen}{1}\color{black}{)}^{2}-2(\color{OliveGreen}{1}\color{black}{)}-1 \\ &=1-2-1 \\ &=-2 \end{aligned}\)

The vertex is \((1,-2)\). We need one more point.

To summarize, we have

\(y\)-intercepts: \((0,1)\)

\(x\)-intercepts: \((1-\sqrt{2}, 0)\) and \((1+\sqrt{2}, 0)\)

Vertex: \((1,-2)\)

Extra point: \((2,-1)\)

Plot the points and sketch the graph.

Answer:

Exercise \(\PageIndex{1}\)

Graph: \(g(x)=-4 x^{2}+12 x-9\).

Answer

www.youtube.com/v/tQiDXNhn7ik

Example \(\PageIndex{4}\):

Determine the maximum or minimum: \(y=-4 x^{2}+24 x-35\).

Solution

Since \(a = −4\), we know that the parabola opens downward and there will be a maximum \(y\)-value. To find it, first find the \(x\)-value of the vertex.

\(\begin{aligned} x &=-\frac{b}{2 a} \quad\:\:\:\quad\color{Cerulean}{x-value\:of\:the\:vertex.} \\ &=-\frac{\color{OliveGreen}{24}}{\color{black}{2}(\color{OliveGreen}{-4}\color{black}{)}} \quad\color{Cerulean}{Substitute\:a=-4\:and\:b-24.} \\ &=-\frac{24}{-8} \quad\quad\:\color{Cerulean}{Simplify.}\\ &=3 \end{aligned}\)

The \(x\)-value of the vertex is \(3\). Substitute this value into the original equation to find the corresponding \(y\)-value.

\(\begin{aligned} y &=-4 x^{2}+24 x-35\quad\quad\quad\color{Cerulean}{Substitute\:x=3.} \\ &=-4(3)^{2}+24(3)-35\quad\:\:\color{Cerulean}{Simplify.} \\ &=-36+72-35 \\ &=1 \end{aligned}\)

The vertex is \((3, 1)\). Therefore, the maximum \(y\)-value is \(1\), which occurs where \(x = 3\), as illustrated below:

Note: The graph is not required to answer this question.

Answer:

The maximum is \(1\).

Example \(\PageIndex{5}\):

Determine the maximum or minimum: \(y=4 x^{2}-32 x+62\).

Solution

Since \(a = 4\), the parabola opens upward and there is a minimum \(y\)-value. Begin by finding the \(x\)-value of the vertex.

\(\begin{aligned} x &=-\frac{b}{2 a} \\ &=-\frac{\color{OliveGreen}{-32}}{\color{black}{2}(\color{OliveGreen}{4}\color{black}{)}} \quad\color{Cerulean}{Substitute\:a=4\:and\:b=-32.}\\ &=-\frac{-32}{8}\:\:\:\:\color{Cerulean}{Simplify.} \\ &=4 \end{aligned}\)

Substitute \(x = 4\) into the original equation to find the corresponding \(y\)-value.

\(\begin{aligned} y &=4 x^{2}-32 x+62 \\ &=4(\color{OliveGreen}{4}\color{black}{)}^{2}-32(\color{OliveGreen}{4}\color{black}{)}+62 \\ &=64-128+62 \\ &=-2 \end{aligned}\)

The vertex is \((4, −2)\). Therefore, the minimum \(y\)-value of \(−2\) occurs where \(x = 4\), as illustrated below:

Answer:

The minimum is \(-2\).

Example \(\PageIndex{6}\):

The height in feet of a projectile is given by the function \(h(t)=-16 t^{2}+72 t\), where \(t\) represents the time in seconds after launch. What is the maximum height reached by the projectile?

Solution

Here \(a = −16\), and the parabola opens downward. Therefore, the \(y\)-value of the vertex determines the maximum height. Begin by finding the time at which the vertex occurs.

\(t=-\frac{b}{2 a}=-\frac{72}{2(-16)}=\frac{72}{32}=\frac{9}{4}\)

The maximum height will occur in \(\frac{9}{4}\) seconds (or \(2 \frac{1}{4}\) seconds). Substitute this time into the function to determine the maximum height attained.

\(\begin{aligned} h\color{black}{\left(\color{OliveGreen}{\frac{9}{4}}\right)} &=-16\color{black}{\left(\color{OliveGreen}{\frac{9}{4}}\right)^{2}}+72\color{black}{\left(\color{OliveGreen}{\frac{9}{4}}\right)} \\ &=-16\left(\frac{81}{16}\right)+72\left(\frac{9}{4}\right) \\ &=-81+162 \\ &=81 \end{aligned}\)

Answer:

The maximum height of the projectile is \(81\) feet.

Example \(\PageIndex{7}\):

Determine the vertex: \(f(x)=2(x+3)^{2}-2\).

Solution

Rewrite the equation as follows before determining \(h\) and \(k\).

\(\begin{array}{c}{f(x)=\:\:\:a\:(\:x\:-\:h\:)^{2}\:\:+\:\:k} \\ \color{Cerulean}{\quad\quad\quad\quad\quad\quad\:\:\:\:\downarrow\quad\quad\:\:\downarrow} \\{f(x)=2[x-(-3)]^{2}+(-2)}\end{array}\)

Here \(h=-3\) and \(k=-2\).

Answer:

The vertex is \((-3,-2)\).

Example \(\PageIndex{8}\):

Rewrite in vertex form and determine the vertex: \(f(x)=x^{2}+4 x+9\).

Solution

Begin by making room for the constant term that completes the square.

\(\begin{aligned} f(x) &=x^{2}+4 x+9 \\ &=x^{2}+4 x+\_\_\_+9-\_\_\_\end{aligned}\)

The idea is to add and subtract the value that completes the square,\(\left(\frac{b}{2}\right)^{2}\), and then factor. In this case, add and subtract \(\left(\frac{4}{2}\right)^{2}=(2)^{2}=4\).

\(\begin{aligned} f(x) &=x^{2}+4 x+9\quad\quad\quad\quad\:\:\color{Cerulean}{Add\:and\:subtract\:4.} \\ &=x^{2}+4 x\color{Cerulean}{+4}\color{black}{+}9\color{Cerulean}{-4}\quad\color{Cerulean}{Factor.} \\ &=\left(x^{2}+4 x+4\right)+5 \\ &=(x+3)(x+2)+5 \\ &=(x+2)^{2}+5 \end{aligned}\)

Adding and subtracting the same value within an expression does not change it. Doing so is equivalent to adding \(0\). Once the equation is in this form, we can easily determine the vertex.

\(\begin{array}{c}{f(x)=a(x-h)^{2}\:\:+\:\:k}\\\color{Cerulean}{\quad\quad\quad\quad\quad\:\:\:\downarrow\quad\:\:\:\:\:\downarrow} \\ {f(x)=(x-(-2))^{2}+5}\end{array}\)

Here \(h=-2\) and \(k=5\).

Answer:

The vertex is \((-2,5)\).

Example \(\PageIndex{9}\):

Rewrite in vertex form and determine the vertex: \(f(x)=2 x^{2}-4 x+8\).

Solution

Since \(a = 2\), factor this out of the first two terms in order to complete the square. Leave room inside the parentheses to add and subtract the value that completes the square.

\(\begin{aligned} f(x) &=2 x^{2}-4 x+8 \\ &=2\left(x^{2}-2 x\right)+8 \end{aligned}\)

Now use \(−2\) to determine the value that completes the square. In this case, \(\left(\frac{-2}{2}\right)^{2}=(-1)^{2}=1\). Add and subtract \(1\) and factor as follows:

\(\begin{aligned} f(x) &=2 x^{2}-4 x+8 \\ &=2\left(x^{2}-2 x+\_\_\_-\_\_\_\right)+8 \quad\color{Cerulean}{Add\:and\:subtract\:1.} \\ &=2\color{black}{\left(x^{2}-2 x\color{Cerulean}{+1-1}\right)}+8\quad\quad\quad\color{Cerulean}{Factor.} \\ &=2[(x-1)(x-1)-1]+8 \\ &=2\left[(x-1)^{2}-1\right]+8 \quad\quad\quad\quad\:\:\color{Cerulean}{Distribute\:the\:2.} \\ &=2(x-1)^{2}-2+8 \\ &=2(x-1)^{2}+6 \end{aligned}\)

In this form, we can easily determine the vertex.

\(\begin{array}{c}{f(x)=a(x-h)^{2}+k}\\\color{Cerulean}{\quad\quad\quad\quad\quad\:\:\downarrow\quad\:\:\:\downarrow} \\ {f(x)=2(x-1)^{2}+6}\end{array}\)

Here \(h=1\) and \(k=6\).

Answer:

The vertex is \((1,6)\).

Exercise \(\PageIndex{1}\)

Rewrite in vertex form and determine the vertex: \(f(x)=-2 x^{2}-12 x+3\).

Answer

\(f(x)=-2(x+3)^{2}+21\); vertex: \((-3,21)\)

www.youtube.com/v/cmwMn6oZ2dI

Exercise \(\PageIndex{3}\)

Does the parabola open upward or downward? Explain.

1. \(y=x^{2}-9 x+20\)
2. \(y=x^{2}-12 x+32\)
3. \(y=-2 x^{2}+5 x+12\)
4. \(y=-6 x^{2}+13 x-6\)
5. \(y=64-x^{2}\)
6. \(y=-3 x+9 x^{2}\)

Answer

1. Upward

3. Downward

5. Downward

Exercise \(\PageIndex{4}\)

Determine the \(x\)- and \(y\)-intercepts.

1. \(y=x^{2}+4 x-12\)
2. \(y=x^{2}-13 x+12\)
3. \(y=2 x^{2}+5 x-3\)
4. \(y=3 x^{2}-4 x-4\)
5. \(y=-5 x^{2}-3 x+2\)
6. \(y=-6 x^{2}+11 x-4\)
7. \(y=4 x^{2}-27\)
8. \(y=9 x^{2}-50\)
9. \(y=x^{2}-x+1\)
10. \(y=x^{2}-6 x+4\)

Answer

1. \(x\)-intercepts: \((-6,0),(2,0)\); \(y\)-intercept: \((0,-12)\)

3. \(x\)-intercepts: \((-3,0),\left(\frac{1}{2}, 0\right)\); \(y\)-intercept: \((0,-3)\)

5. \(x\)-intercepts: \((-1,0),\left(\frac{2}{5}, 0\right)\); \(y\)-intercept: \((0,2)\)

7. \(x\)-intercepts: \(\left(-\frac{3 \sqrt{3}}{2}, 0\right),\left(\frac{3 \sqrt{3}}{2}, 0\right)\); \(y\)-intercept: \((0,-27)\)

9. \(x\)-intercepts: none; \(y\)-intercept: \((0,1)\)

Exercise \(\PageIndex{5}\)

Find the vertex and the line of symmetry.

1. \(y=-x^{2}+10 x-34\)
2. \(y=-x^{2}-6 x+1\)
3. \(y=-4 x^{2}+12 x-7\)
4. \(y=-9 x^{2}+6 x+2\)
5. \(y=4 x^{2}-1\)
6. \(y=x^{2}-16\)

Answer

1. Vertex: \((5,-9)\); line of symmetry: \(x=5\)

3. Vertex: \(\left(\frac{3}{2}, 2\right)\); line of symmetry: \(x=\frac{3}{2}\)

5. Vertex: \((0,-1)\); line of symmetry: \(x=0\)

Exercise \(\PageIndex{6}\)

Graph. Find the vertex and the \(y\)-intercept. In addition, find the \(x\)-intercepts if they exist.

1. \(f(x)=x^{2}-2 x-8\)
2. \(f(x)=x^{2}-4 x-5\)
3. \(f(x)=-x^{2}+4 x+12\)
4. \(f(x)=-x^{2}-2 x+15\)
5. \(f(x)=x^{2}-10 x\)
6. \(f(x)=x^{2}+8 x\)
7. \(f(x)=x^{2}-9\)
8. \(f(x)=x^{2}-25\)
9. \(f(x)=1-x^{2}\)
10. \(f(x)=4-x^{2}\)
11. \(f(x)=x^{2}-2 x+1\)
12. \(f(x)=x^{2}+4 x+4\)
13. \(f(x)=-4 x^{2}+12 x-9\)
14. \(f(x)=-4 x^{2}-4 x+3\)
15. \(f(x)=x^{2}-2\)
16. \(f(x)=x^{2}-3\)
17. \(f(x)=-4 x^{2}+4 x-3\)
18. \(f(x)=4 x^{2}+4 x+3\)
19. \(f(x)=x^{2}-2 x-2\)
20. \(f(x)=x^{2}-6 x+6\)
21. \(f(x)=-2 x^{2}+6 x-3\)
22. \(f(x)=-4 x^{2}+4 x+1\)
23. \(f(x)=x^{2}+3 x+4\)
24. \(f(x)=-x^{2}+3 x-4\)
25. \(f(x)=-2 x^{2}+3\)
26. \(f(x)=-2 x^{2}-1\)
27. \(f(x)=2 x^{2}+4 x-3\)
28. \(f(x)=3 x^{2}+2 x-2\)

Answer

1.

3.

5.

7.

9.

11.

13.

15.

17.

19.

21.

23.

25.

27.

Exercise \(\PageIndex{7}\)

Determine the maximum or minimum \(y\)-value.

1. \(y=-x^{2}-6 x+1\)
2. \(y=-x^{2}-4 x+8\)
3. \(y=25 x^{2}-10 x+5\)
4. \(y=16 x^{2}-24 x+7\)
5. \(y=-x^{2}\)
6. \(y=1-9 x^{2}\)
7. \(y=20 x-10 x^{2}\)
8. \(y=12 x+4 x^{2}\)
9. \(y=3 x^{2}-4 x-2\)
10. \(y=6 x^{2}-8 x+5\)
11. \(y=x^{2}-5 x+1\)
12. \(y=1-x-x^{2}\)

Answer

1. Maximum: \(y = 10\)

3. Minimum: \(y = \frac{4}{5}\)

5. Maximum: \(y = 0\)

7. Maximum: \(y = 10\)

9. Minimum: \(y = −\frac{10}{3}\)

11. Minimum: \(y = −\frac{21}{4}\)

Exercise \(\PageIndex{8}\)

Given the following quadratic functions, determine the domain and range.

1. \(f(x)=3 x^{2}+30 x+50\)
2. \(f(x)=5 x^{2}-10 x+1\)
3. \(g(x)=-2 x^{2}+4 x+1\)
4. \(g(x)=-7 x^{2}-14 x-9\)
5. \(f(x)=x^{2}+x-1\)
6. \(f(x)=-x^{2}+3 x-2\)
7. The height in feet reached by a baseball tossed upward at a speed of \(48\) feet per second from the ground is given by the function \(h(t) = −16t^{2} + 48t\), where \(t\) represents the time in seconds after the ball is thrown. What is the baseball’s maximum height and how long does it take to attain that height?
8. The height in feet of a projectile launched straight up from a mound is given by the function \(h(t) = −16t^{2} + 96t + 4\), where \(t\) represents seconds after launch. What is the maximum height?
9. The profit in dollars generated by producing and selling \(x\) custom lamps is given by the function \(P(x) = −10x^{2} + 800x − 12,000\). What is the maximum profit?
10. The profit in dollars generated from producing and selling a particular item is modeled by the formula \(P(x) = 100x − 0.0025x^{2}\), where \(x\) represents the number of units produced and sold. What number of units must be produced and sold to maximize revenue?
11. The average number of hits to a radio station Web site is modeled by the formula \(f(x) = 450t^{2} − 3,600t + 8,000\), where \(t\) represents the number of hours since \(8:00\) a.m. At what hour of the day is the number of hits to the Web site at a minimum?
12. The value in dollars of a new car is modeled by the formula \(V(t) = 125t^{2} − 3,000t + 22,000\), where \(t\) represents the number of years since it was purchased. Determine the minimum value of the car.
13. The daily production cost in dollars of a textile manufacturing company producing custom uniforms is modeled by the formula \(C(x) = 0.02x^{2} − 20x + 10,000\), where \(x\) represents the number of uniforms produced.
    1. How many uniforms should be produced to minimize the daily production costs?
    2. What is the minimum daily production cost?
14. The area in square feet of a certain rectangular pen is given by the formula \(A = 14w − w^{2}\), where \(w\) represents the width in feet. Determine the width that produces the maximum area.

Answer

1. Domain: \((−∞, ∞)\); range:\([−25, ∞)\)

3. Domain: \((−∞, ∞)\); range: \((−∞, 3]\)

5. Domain: \((−∞, ∞)\); range: \([−\frac{5}{4}, ∞)\)

7. The maximum height of \(36\) feet occurs after \(1.5\) seconds.

9. $\(4,000\)

11. \(12:00\) p.m.

13. (1) \(500\) uniforms (2) $\(5,000\)

Exercise \(\PageIndex{9}\)

Determine the vertex.

1. \(y=-(x-5)^{2}+3\)
2. \(y=-2(x-1)^{2}+7\)
3. \(y=5(x+1)^{2}+6\)
4. \(y=3(x+4)^{2}+10\)
5. \(y=-5(x+8)^{2}-1\)
6. \(y=(x+2)^{2}-5\)

Answer

1. \((5, 3)\)

3. \((−1, 6)\)

5. \((−8, −1)\)

Exercise \(\PageIndex{10}\)

Rewrite in vertex form \(y=a(x-h)^{2}+k\) and determine the vertex.

1. \(y=x^{2}-14 x+24\)
2. \(y=x^{2}-12 x+40\)
3. \(y=x^{2}+4 x-12\)
4. \(y=x^{2}+6 x-1\)
5. \(y=2 x^{2}-12 x-3\)
6. \(y=3 x^{2}-6 x+5\)
7. \(y=-x^{2}+16 x+17\)
8. \(y=-x^{2}+10 x\)

Answer

1. \(y=(x-7)^{2}-25\); vertex: \((7, -25)\)

3. \(y=(x+2)^{2}-16\); vertex: \((-2, -16)\)

5. \(y=2(x-3)^{2}-21\); vertex: \((3, -21)\)

7. \(y=-(x-8)^{2}+81\); vertex: \((8, 81)\)

Exercise \(\PageIndex{11}\)

Graph. Find the vertex and the \(y\)-intercept. In addition, find the \(x\)-intercepts if they exist.

1. \(f(x)=x^{2}-1\)
2. \(f(x)=x^{2}+1\)
3. \(f(x)=(x-1)^{2}\)
4. \(f(x)=(x+1)^{2}\)
5. \(f(x)=(x-4)^{2}-9\)
6. \(f(x)=(x-1)^{2}-4\)
7. \(f(x)=-2(x+1)^{2}+8\)
8. \(f(x)=-3(x+2)^{2}+12\)
9. \(f(x)=-5(x-1)^{2}\)
10. \(f(x)=-(x+2)^{2}\)
11. \(f(x)=-4(x-1)^{2}-2\)
12. \(f(x)=9(x+1)^{2}+2\)
13. \(f(x)=(x+5)^{2}-15\)
14. \(f(x)=2(x-5)^{2}-3\)
15. \(f(x)=-2(x-4)^{2}+22\)
16. \(f(x)=2(x+3)^{2}-13\)

Answer

1.

3.

5.

7.

9.

11.

13.

15.

Exercise \(\PageIndex{12}\)

1. Write down your plan for graphing a parabola on an exam. What will you be looking for and how will you present your answer? Share your plan on the discussion board.
2. Why is any parabola that opens upward or downward a function? Explain to a classmate how to determine the domain and range.
3. Research and discuss ways of finding a quadratic function that has a graph passing through any three given points. Share a list of steps as well as an example of how to do this.

Answer

1. Answer may vary

3. Answer may vary