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# I need some help with these example parabola problems

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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.

Feb 9, 2019

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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)   Feb 9, 2019
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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)   Feb 9, 2019
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just need help with 4 & 5 please? Feb 9, 2019
#4
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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
edited by CPhill  Feb 9, 2019
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4) the other x-intercept would be the same distance from the axis of symetry as the x-intercept that they have given you.

If it is 3 places over than the other one would be (-3) places from the axis of symetry.

Feb 9, 2019