Posted by **-Untamed-** on Monday, December 5, 2011 at 8:08pm.

The graph of a quadratic function has vertex(3,-4) and passes through the point (4,1).

Find the equation of the function in standard form.

Rewrite the equation in general form.

Could someone explain to me how to do this, just don't understand.

- Math -
**MathMate**, Monday, December 5, 2011 at 9:00pm
The canonical form of a quadratic equation is:

y=a(x-h)²+k

where

a is a constant

(h,k) is the vertex.

So in the given case, h=3, k=-4, and a remains to be found.

We can find a since we know the curves passes through (4,1), hence

1=a(4-3)²-4

1=a-4

solve for a to get

a=5

Check:

y=5(x-3)²-4

For x=4, y=5(1)²-4=1, OK

For the general form, expand the equation

y=5(x-3)²-4

y=5x²-30x+41

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