Question

f(x)=4x^2-7x-15 how do i factor it and find the x-intercepts? Also please describe the end behavior of the graph of f(x) and the steps you would use to graph. Pleaseee im really lost

Answer 1

Given:

The function is

`f(x)=4x^2-7x-15`

Required:

Find the x-intercepts and describe the end behavior of the graph of f(x).

Step-by-step explanation:

The given function is:

`f(x)=4x^(2)-7x-15`

To find the x-intercepts put f(x)=0

`4x^2-7x-15=0`

This is a quadratic equation solved by using the middle term splitting method.

`\begin{gathered} 4x^2-12x+5x-15=0 \\ 4x(x-3)+5(x-3)=0 \\ (x-3)(4x+5)=0 \end{gathered}`
`\begin{gathered} x-3=0 \\ x=3 \end{gathered}`
`\begin{gathered} 4x+5=0 \\ x=-(5)/(4) \\ x=-1.25 \end{gathered}`

Thus the x-intercept are: -1.25 and 3.

Compare the given equation with the standard equation

`y=ax^2+bx+c`
`a=4,b=-7,c=-15`

The vertex is (h,k).

`\begin{gathered} h=-(b)/(2a) \\ h=-(-7)/(2*4) \\ h=(7)/(8) \\ h\approx0.9 \end{gathered}`
`\begin{gathered} k=f(h) \\ k=f(0.9) \\ k=4(0.9)^2-7(0.9)-15 \\ k=3.24-6.3-15 \\ k\approx-18.1 \end{gathered}`

Thus the vertex of the parabola is:

`(0.9,-18.1)`

We will plot the points of the vertex and the x-intercepts and then draw the graph of the parabola as:

Here the leading coefficient a = 4>0

So the parabola opens upwards.

In the given function the degree is 2 which is even so as

Final Answer:

As explained in the explanation part.