Question

Could any one help me figure out the last two parts? Thanks

Answer 1

Given the equation of a quadratic function:

`f(x)=x^2-3x-28`

First, we will find the x-intercepts to find the largest x-intercept

So, substitute f = 0, then solve for x as follows:

`\begin{gathered} x^2-3x-28=0 \\ (x-7)(x+4)=0 \\ x-7=0\rightarrow x=7 \\ x+4=0\rightarrow x=-4 \end{gathered}`

So, The largest x-intercept = 7

Second, we will find the y-coordinate of the y-intercept

So, substitute x = 0

`f(x)=(0)^2-3(0)-28=-28`

So, The y-coordinate of the y-intercept = -28