Question

a parabola has the given x-intercepts. what is the x coordinate of the vertex? write your result in decimal form.
(-4,0) (3,0)

Answer 1

Equation of the parabola: \[ y=(x+4)(x-3)=x^2+x-12 \]

Vertex is at \(\left(-\frac 1 2, -\frac {49}{4}\right) \)

Answer 2

Since it has the x-ints -4, 3 , that means the solution is -4, 3
Then you can form a quadratic equation with the given roots as
[x-(-4)](x-3) =0
(x+4)(x-3)=0
Expand it, you'll get
x^2 + 4x -3x -12 =0
x^2 +x -12 =0
Now, using the method of completing square.
(x+x+ (1/2)^2) - (1/2)^2 -12 =0
(x+1/2)^2 - 49/4 =0
Since vertex is (h,k) in the equation y=a(x-h)^2+k
Compare it and you will get h=-1/2, k=-49/4
So, the vertex is (-1/2 , -49/4)