+76 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.
+129852 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)
+129852 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)
+129852 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
