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YALL I NEED HELP ASAP Use the function f(x) to answer the questions.

f(x) = −16x2 + 24x + 16

Part A: What are the x-intercepts of the graph of f(x)? Show your work. (2 points)

Part B: Is the vertex of the graph of f(x) going to be a maximum or a minimum? What are the coordinates of the vertex? Justify your answers and show your work. (3 points)

Part C: What are the steps you would use to graph f(x)? Justify that you can use the answers obtained in Part A and Part B to draw the graph. (5 points)

Sagot :

Answer:

See the below works.

Step-by-step explanation:

We can graph the function f(x) = -16x² + 24x + 16 by these steps:

  • find the x-intercepts, where the value of f(x) = 0
  • find the y-intercept, where the value of x = 0
  • determine the direction of the opening and find the vertex
  • (optional) find 2 additional coordinates that are reasonably far from the left and right of the above points to make a better graph

Part A:

When the graph intersects the x-axis, the y-value (f(x)) equals to 0. Therefore, to find the x-intercepts, we substitute f(x) with 0:

[tex]\begin{aligned}f(x)&=-16x^2+24x+16\\0&=-16x^2+24x+16\\0&=2x^2-3x-2\\0&=(2x+1)(x-2)\\x&=-\frac{1}{2} \ or\ 2\end{aligned}[/tex]

[tex]\texttt{Hence, the coordinates }=(-\frac{1}{2} ,0)\ \texttt{and}\ (2,0)[/tex]

Part B:

To determine the opening direction of the graph (quadratic function), we apply these rules:

  • If the coefficient of x² < 0 ⇒ the graph opens downwards and the vertex is a maximum
  • If the coefficient of x² > 0 ⇒ the graph opens upwards and the vertex is a minimum

For this question, the coefficient of x² is -16, which is smaller than 0, then the graph opens downwards and the vertex is a maximum.

To find the vertex, we use this formula:

[tex]\boxed{(x,y)=\left(-\frac{b}{2a} ,\frac{-(b^2-4ac)}{4a} \right)}[/tex]

[tex]\begin{aligned}(x,y)&=\left(-\frac{24}{2(-16)} ,\frac{-(24^2-4(-16)(16))}{4(-16)} \right)\\\\&=\left(\frac{3}{4} ,25\right)\end{aligned}[/tex]

Part C:

Refer to the top part for the steps to graph f(x). From Part A and B, we have the x-intercepts and the vertex. Now, we need to find the y-intercept. When the graph intersect the y-axis, the x-value equals to 0. Therefore, to find the y-intercepts, we substitute x with 0:

[tex]\begin{aligned}f(x)&=-16x^2+24x+16\\&=-16(0)^2+24(0)+16\\&=16\end{aligned}[/tex]

[tex]\texttt{Hence, the coordinates }=(0,16)[/tex]

Now, we have all 4 coordinates needed and we can connect them to create the graph. (see the attached drawing)

Note:

To make a nicer graph, we can find 2 coordinates that are reasonably far from the known coordinates. Since the most left point is [tex](-\frac{1}{2} ,0)[/tex] and the most right point is [tex](2,0)[/tex], then we can find coordinates with x = -1 and 3 to make a nicer graph.

[tex]f(-1)=-16(-1)^2+24(-1)+16=-24[/tex]

[tex]f(3)=-16(3)^2+24(3)+16=-56[/tex]

Then by connecting with points (-1, -24) and (3, -56) will make the graph looks nicer.

View image Karmenchong
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