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

User Metsburg
by
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2 Answers

4 votes
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
User Mtkale
by
7.4k points
0 votes
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
User Danielgatis
by
8.6k points

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