Question

I need some help with graphing quadratics.
How do I find the x and y-intercepts in both y=(x-h)^2+k and y=ax^2+bx+c forms?

Answer 1

for x intercept, assume y=0 and for y intercept, assume x= 0........substitute these value in equation and you will get answer

Answer 2

y=ax^2+bx+c

solve the quadratic equation to get the roots (x-intercepts)


when y intercept put x = 0
and solve.

Answer 3

y=(x-h)^2+k
this form is basically used to find thevertex.

Answer 4

the vertex*

Answer 5

y=(x-1)^2-16
By substituting x for zero, I got -15. So the y-intercept is 0,-15.
I'm still confused on how to find the x-intercept.

Answer 6

correct.

now,

y=(x-1)^2-16
(x-1)(x-1) - 16 = 0

solve this using foil.

Answer 7

x^2-2x-15

Answer 8

correct.

now use the factorization or use the quadratic formula to solve this quadratic equation.

Answer 9

3,0 and -5,0

Answer 10

wouldn't it be,

-3,0 and 5,0

Answer 11

No, because when you add -3 and 5 you get 2, not -2.

Answer 12

x^2-2x-15 =

x^2 - 5x + 3x - 15 = 0

x(x-5) + 3(x-5) = 0

(x+3)(x-5) = 0

(x+3) = 0 | (x+5) = 0
x = -3 | x = -5