Question

suppose a parabola has vertex (5, –3) and also passes through the point (6, 1). Write the equation of the parabola in vertex form.
y = (x – 5)2 – 3
y = 4(x – 5)2 – 3
y = 4(x + 5)2 – 3
y = 4(x – 5)2 + 3

Answer 1

Since y = x^2 is the only form offered, we can ignore the x = y^2 variety.

Should be an immediate result this far:

(y-(-3)) = (something)(x-5)^2

This discards #3. It has "x+5"

Standard form

(y+3) = (something)(x-5)^2
y = (something)(x-5)^2 - 3

This discards #4. It has "+3".

Substitute (6,1)

1 = (something)(6-5)^2 - 3
1 = (something)(1)^2 - 3
1 = (something) - 3
Something = 4

Discard #1 and we're done.

Really, just one step at a time. Walk through it in a way that is logical and organized. No need to see the solution from the beginning, Just take a step.

Answer 2

so its y = 4(x – 5)2 – 3

Answer 3

Please use the caret for exponentiation. x2 is very odd-looking. x^2 means x-squared.

You tell me. Is it?