Question 199466
Manipulate your equation into the "vertex form":
y= a(x-h)^2+k 
where
(h,k) is the vertex
.
f(x) = 3x^2 + 6x - 7
f(x) = (3x^2 + 6x) - 7
f(x) = 3(x^2 + 2x) - 7
Complete the square:
f(x) = 3(x^2 + 2x + 1) - 7 - 3
f(x) = 3(x+1)^2 - 10
.
Therefore, the vertex is:
(h,k) = (-1, -10) 
.
To find the y-intercepts, set x=0 and solve for f(x):
f(x) = 3x^2 + 6x - 7
f(x) = 3(0^2) + 6(0) - 7
f(x) = - 7
y-intercept is (0, -7)
.
To find x-intercepts, set f(x)=0 and solve for x:
0 = 3x^2 + 6x - 7
Using the quadratic equation, we get:
x = {0.826, -2.826}
x-intercepts are at:
(-2.826,0) and (0.826,0)
.
Details of quadratic follows:
*[invoke quadratic "x", 3, 6, -7 ]