Question

Function = y=(x-4)^2-5+7

Answer 1

Given equation of parabola is $y=(x-4)^2-5+7$

Or we can simplify that to $y=(x-4)^2+2$

This equaion looks similar to formula $y=a(x-h)^2+k$

Comparing both equation, we get:

a=1, h=4, k=2

(h,k) represents the vertex of the parabola

Hence vertex of the given parabola is (4,2).

So x-value of the vertex = 4

So y-value of the vertex = 2

Axis of symmetry is given by equation x=h

So the answer will be x=4

Now we will set y=0 and solve this using quadratic formula including desriminant

$0=(x-4)^2+2$

$0=x^2-8x+16+2$

$0=x^2-8x+18$

compare with quadratic equation  $0=ax^2+bx+c$, we get:

a=1, b=-8, c=18

Descriminant is given by formula:

$Descriminant = \sqrt{b^2-4ac} = \sqrt{(-8)^2-4(1)(18)}= \sqrt{-8}= -2\sqrt{2} i$

which is imaginary

Hence there will be no real solution for x-intercept.