The standard form of a quadratic is
y = a(x-h)^2 + k, where (h,k) is the vertex.
since we know the vertex we can start with
y = a(x-2)^2 + 5 , leaving us only with the a as an unknown, but we know the point (-8,15) lies on it, thus must satisfy our equation, so ...
15 = a(-8-2)^2 + 5
10 = 100a
a = 10/100 = 1/10
equation: y = (1/10)(x-2)^2 + 5
2.
for the y-intercept, let x = 0
we can visually see the would give us y = 6
for the x-intercept, let y = 0
then
2x^2 + x - 6 = 0
(2x - 3)(x + 2) = 0
x = 3/2 or x = -2
1. Find the equation for a quadratic function whose vertex is (2,5) and whose graph contains the point (-8,15).
2.What are the x- and y- intercepts for y=2x^2+x-6
Plesee explai. I want to learn to learn how to do them.
1 answer