1. Find the vertex of the graph of y=x^2+8x+23

2.Find the vertex of the graph of the equation y=3x^2-6x+7

3.Find the equation of the axis of symmetry of the graph of y= (x+3)^2-5

4.The vertex of a parabola is at (12,-4), and its axis of symmetry is vertical. One of the x-intercepts is at . What is the x-coordinate of the other x-intercept?

5.The x-intercepts of the parabola y=x^2+bx+c are (-3,0) and (5,0) Find the equation of the parabola, and submit your answer in y=ax^2+bx+c form.

Android4EVER Feb 9, 2019

#1**+2 **

1. Find the vertex of the graph of y=x^2+8x+23

In the form y = ax^2 + bx + c.....the x coordinate of the vertex is given by -b / [ 2a ]

So

The x coordinate of the vertex is -8/ [ 2 * 1 ] = -8/2 = -4

And the y coordinte of the vertex is y = (-4)^2 + 8(-4) + 23 = 16 - 32 + 23 = 7

So....the vertex = (-4, 7)

CPhill Feb 9, 2019

#2**+2 **

2.Find the vertex of the graph of the equation y=3x^2-6x+7

Similar to the last one

x cooordinate of the vertex is - [ -6] / [2 * 3 ] = 6 / 6 = 1

And the y coordinate of the vertex is y = 3(1)^2 - 6(1) + 7 = 3 - 6 + 7 = 4

So...the vertex is (-1, 4)

CPhill Feb 9, 2019

#3

#4**+1 **

Got it, Android...

4. I need to know what the first x-intercept is.....

5. The x-intercepts of the parabola y=x^2+bx+c are (-3,0) and (5,0) Find the equation of the parabola, and submit your answer in y=ax^2+bx+c form.

If (-3, 0) and (5, 0) are x intercepts...then they are also roots

So we have that

(x - - 3) ( x - 5) = 0

(x + 3) ( x - 5) = 0 expand and simplify

x^2 - 2x - 15 = 0

So....the equation becomes

y = x^2 - 2x - 15

CPhill
Feb 9, 2019