Question

Find the x-intercepts of the parabola with vertex (6,-5) and y-intercept (0,175) .

Answer 1

Vertex= (6,-5)
y-intercept=(0,175)

(x-6)^2-5=0
(x-6)^2=5
x-6=+- square root 5
x=6+- square root 5
x=8.24, x=3.76
x-intercepts= 8.24 and 3.76

Answer 2

* note * i just noticed i answered for the wrong thing, however i will leave this in case someone wants to learn it

First you would plug in the values of the y-intercept into a generic quadratic equation. A generic quadratic equation is:
y = ax^2 + bx + c

so plugging in the y intercept values of (0, 175) we get
175 = 0x^2 + 0b + c

this shows what c equals since 0 times any number is zero:
c = 175

now the generic vertex form is written as:
y - k = a * (x - h)^2 or
y = a * (x - h)^2 + k

with (h, k ) being the vertex coordinates

I will use 2nd equation and plug in vertex point
y = a * (x - 6)^2 -5
y = a* (x^2 - 12x + 36) - 5
y = ax^2 - 12x + 36a - 5

Since we know that the standard equation is
y = ax^2 +bx + c

we can use what we just found and plug it into the standard equation but only using 1 of its 3 parts at a time (the parts being ax^2, bx, and c). So:
ax^2 = ax^2, this does nothing

bx = -12x, this can be reduced by dividing both sides by x
b = -12

c = 36a - 5, from earlier we know that c = 175 so
175 = 36a -5
180 = 36a
a = 5

now we have all three values and can plug in to get our equation
y = 5x^2 - 12x + 175