Question

If your able to answer all of them i will be giving you 5 stars

Answer 1

The function given is:

`f(x)=-16x^2+60x+16`PART A

The factorization steps are shown below:

`\begin{gathered} f(x)=-16x^2+60x+16 \\ f(x)=4(-4x^2+15x+4) \\ f(x)=4(-4x^2+16x-x+4) \\ f(x)=4(-4x(x-4)-1(x-4)) \\ f(x)=4(-4x-1)(x-4) \end{gathered}`PART B

To find the x intercepts, we set f(x) equal to 0 and solve for x:

`\begin{gathered} f(x)=4(-4x-1)(x-4) \\ f(x)=0 \\ 4(-4x-1)(x-4)=0 \\ -4x-1=0--------(1) \\ OR \\ x-4=0---------(2) \end{gathered}`

Solving (1), we have:

`\begin{gathered} -4x-1=0 \\ 4x=-1 \\ x=-(1)/(4) \\ x=-0.25 \end{gathered}`

and, solving (2), we have:

`\begin{gathered} x-4=0 \\ x=4 \end{gathered}`

The x-intercepts are

`\begin{gathered} x=-0.25 \\ x=4 \end{gathered}`

PART C

The standard equation of a quadratic is

`f(x)=ax^2+bx+c`

The parabola opens upward when a is positive and opens downward when a is negative

1. When parabola opens upward, the end behavior can be described as:

`\begin{gathered} x\rightarrow\infty \\ y\rightarrow\infty \\ \text{and} \\ x\rightarrow-\infty \\ y\rightarrow\infty \end{gathered}`

2. When parabola opens downward, the end behavior can be described as:

`\begin{gathered} x\rightarrow\infty \\ y\rightarrow-\infty \\ \text{and} \\ x\rightarrow-\infty \\ y\rightarrow-\infty \end{gathered}`

Our equation has an "a" value that is negative! So, the parabola opens downward and the end behvaior can be described as:

As x goes to infinity (gets infinitely large), y goes to negative infinity (gets infinitely small) and as x goes to negative infinity (gets infinitely small), y goes to negative infinity (get infinitely small).

PART D

In Part B, we found the x-intercepts. Those are the x-axis cutting points. We can draw those first.

Then,

Using the end behavior information that we found in Part C, we can draw the parabola. The rough sketch is shown:

The exact graph is shown below, for reference: