Question

Find the X-intercepts of the parabola with vertex (1, -9) and y-intercept at (0,-6)

Answer 1

Formula of the parabola:  (x - h)² = 4p(y - k)
(h,k) is a vertex, (1,-9) is a vertex.
h=1, k= - 9

Substitute h and k into formula (x - h)² = 4p(y - k)
(x - 1)² = 4p(y + 9)

Substitute x and y of the y-intercept
(x - 1)² = 4p(y + 9), x=0,y=-6.
(0 - 1)² = 4p(-6 + 9)
1 = 4p*3
1/3 = 4p, substitute value of 4p into (x - 1)² = 4p(y + 9).
 
(x - 1)² = 1/3(y + 9)
x²-2x+1=(1/3)y + 3
(1/3)y = x²-2x-2
y = 3x² - 6x -6

x-intercepts, is values of x when y=0.
3x² - 6x -6 = 0
x²-2x-2=0

We will find values of x using formula.
$x= \frac{-b+/- \sqrt{b^{2}-4ac} }{2a} \\ \\ a=1, b=-2,c=-2 \\ \\x= \frac{2+/- \sqrt{(-2)^{2}-4*1*(-2)} }{2*1} \\ \\x= \frac{2+/- \sqrt{4+8} }{2} \\ \\x= \frac{2+/- \sqrt{12} }{2} = \frac{2+/- 2\sqrt{3} }{2}=1+/- \sqrt{3} \\ \\x_{1} =1- \sqrt{3} , x_{2} =1+ \sqrt{3} \\ \\x_{1} =-0.732, x_{2} =2.732$