Question

Find the x-intercept of the parabola with vertex (1,-13) and y-intercept (0,-11).

Answer 1

1. find equation

general solutions are:

`y = a x^(2) +bx +c \\ \\ (dy)/(dx) = 2ax +b`

for P(0,-11)


`-11 = 0a + 0b +c` ⇒
`c = -11`

for p(1,-13)

`-13 = 1a + 1b -11 \\ a+b = -2`

and


`0 = 2a + b \\ b = -2a`

solve for a and b:

`a - 2a = -2 \\ a = 2 \\ b = -4`

the total equation is now:


`y = 2 x^(2) -4x -11`

To find the x-intercept set y=0 and solve for x


`0 = 2 x^(2) -4x-11`

Answer 2

Standard form of equation for a parabola: y=A(x-h)^2+k, (h,k)=(x,y) coordinates of vertex
using given coordinates of vertex:
y=A(x-1)^2-13
using given coordinates of y-intercept:
-11=A(0-1)^2-13
A=2
Equation of parabola: y=2(x-1)^2-13
x-intercepts

set y=0
0=2(x-1)^2-13
2(x-1)^2=13
(x-1)^2=13/2
x-1=±√(13/2)
x=1±√(13