Find possible formulas for quadratic functions

Question

(a) Graph has vertex (−1, 3) and y-intercept 8.

(b) Graph has x-intercepts −1, 5 and y-intercept −15.

Answer 1

(-1,3) vertex

y=a(x+1)^2 +3 has vertex (-1,3). y=a(x-h)^2 + k has vertex (h,k)

to solve for a, plug in the point (0,8)

8=a+3

y=(5)(x+1)^2 +3 has vertex (-1,3) and y intercept 8

that's an upward opening parabola and the quadratic you probably want

but you could also have a rightward opening parabola with the same vertex and y intercept

x=a(y-3)^2 -1

0=a(8-3)^2 -1 = 25a-1

25a = 1

a =1/25

x=(1/25)(y-3)^2 -1

There're an infinite number of possible parabolas through those two points

x-intercepts -1, and 5, and y intercept -15

y=a(x+1)(x-5) has x intercepts -1 and 5

plug in the y intercept to solve for a

-15 = a(1)(-5) =-5a

a =3

y=3(x+1)(x-5) is the quadratic, a rightward opening parabola with the given intercepts

or multiply it out

y=3(x^2-4x-5) or complete the square adding 12 and subtracting 12

y=3(x^2-4x +4) -15 -12

y = 3(x-2)^2 -27 in vertex form with vertex = (2,-27)

Answer 2

(a) When they give you the vertex, use the vertex form:

y = a·(x-h)² + k

where a is a constant and (h,k) is the location of the vertex. In your case, (h,k) - (-1,3) so the equation so far is:

y = a·(x-(1-))² + 3 = a·(x+1)² + 3

To find the value of the constant a, plug in the given point (y-intercept), (0,8):

8 = a·(0+1)² + 3

8 = a·1 + 3 = a + 3

5 = a

y = 5·(x+1)² + 3

(b) When they give you the x-intercepts, use the factored form:

y = a·(x-p)(x-q)

where a is a constant and p and q are the intercepts. In your case, p = -1 and q = 5:

y = a·(x-(-1))(x-5) = a·(x+1)(x-5)

To find the value of the constant a, plug in the given point (0,-15) and solve for a like we did in the first problem.

Answer 3

First solve for "a" in the vertex formula.

y=a(x-h)²+k This is the vertex formula where (h,k)=(-1,3) is the vertex

8=a(0+1)²+3 Plug in the given numbers (-1,3) and (0,8) y-intercept always has its partner x as zero.

So, a=5

Then rewrite the formula with only x and y appearing and all other variables plugged in.

Final form is y=5(x+1)²+3