 # Math - Vertex Form

In this lecture we were introduced to the vertex form, or squared form, of the quadratic equation.

This form is incredibly useful for quickly seeing where the vertex of the parabola is - hence it’s name.

For your challenge, find the equation in vertex form for the curve that runs through the points (2,0), (5,3), and (8,0).

In intercept form, this was y = -1/3(x - 2)(x - 8)

Remember that the answer will be in the form; y = a(x - h)^2 + k.

For a hint: You can take the ‘a’ from the other form of the equation.

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y = a (x - h)² + k

since a = -(1/3)
h is the vertex x, so 5.
k is the vertex y, so 3.

y = -(1/3) (x - 5)² + 3

and to verify

y = -(1/3) (x - 5)(x - 5) + 3
y = -(1/3) (x² -10x + 25) +3
y = -(1/3)x² + (10/3)x - (25/3) + (9/3)
y = -(1/3)x² + (10/3)x - (16/3)

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I got:

y = -1/3(x-5) + 3

Which converted back to standard form is

y = -1/3x + 10/3x - 16/3

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@Ajai_Raj, you forgot to square the (x-5) but other than that you’re spot on 1 Like

y=−(1/3) (x−5)^2+3
y=−(1/3)(x−5)(x−5)+3
y=−(1/3)(x^2−5x−5x+25)+3
y=−(1/3)(x^2−10x+25)+3
y=−(1/3) x^2−(10/3)x−(25/3)+(9/3)
y=−(1/3) x^2−(10/3)x−(16/3)

For context this is what it looks like in onenote (which I’m pasting from) - the formatting doesn’t seem to come over. 1 Like

It took me a while to get right because I added the constant k before dividing (x-h)2 by a

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