Please please help me with this question please now and show your steps

Y = x^2 - 8x + 7

a) y-intercept

The y-intercept is the point where it crosses the y axis so x would 0.
So putting in 0 for x:
y = x^2 - 8x + 7
y = 0^2 - 8(0) + 7
y = 0 - 0 + 7
y = 7
The y intercept is at the point (0,7).

b)
The zeroes are the points where it crosses the x axis so it’s when the y is set to 0.

0 = x^2 - 8x + 7
Since it’s a quadratic equation we have to factor it. We fill in 2 numbers that add to -8 and multiplied to equal 7.
(x +)(x+ )

The 2 numbers are 7 and -1 because -1*-7=7 and -7+(-1)=-8. So:
0 = (x-7)(x-1)

So now we have x-7 and x-1 equaling some quantity. If either of those happens to equal zero, it makes the whole equation 0, because (x-7)0=0 and 0(x-1)=0

So we can set each of it separately to 0 to solve for x.
x-7=0
x= 7

x-1=0
x= 1

The zeroes are at the points (1, 0) and (7, 0)

C.
The Vertex Form of a Quadratic is:
y = a(x-h)^2 + k. And the vertex is (h, k)

y = x^2 - 8x + 7
y - 7 = x^2 - 8x

We want to complete the square by adding some number to both sides of the equation so that when we factor it we are going to get (x-a)(x-a).

So take half of 8 to get two equal numbers that would be our “a”. 8/2=4 so we want (x - 4)(x- 4).

(x-4)(x-4) multiplies out to x^2 - 8 + 16.

continuing,
y - 7 = x^2 - 8x

Add 16 to both sides.
y - 7 + 16 = x^2 - 8x + 16

y + 9 = (x-4)(x-4)
y + 9 = (x-4)^2
y = (x-4)^2 - 9 is the vertex form.

The vertex is at the point (4, -9). So the axis of symmetry is at the line x = 4.

D. The vertex is solved in C as the point (-4, -9).

No it doesn’t because in a function you can’t use a number twice on an input or output