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November 28, 2014

November 28, 2014

Posted by **Raquel** on Thursday, June 25, 2009 at 10:35am.

y=x^2-3x-12 and y=2x-16

and epxress y=x^2-3x-12 in the form y=(x-h^2)+k, where h and k are constants

Hence determine the minimum value of the function y=x^2-3x-12

I have a few of these to do, so if someone can just show me how to do this one i can do the rest.

- Mathematics -
**Reiny**, Thursday, June 25, 2009 at 10:53amin the first:

notice that one equation is quadratic, the other is linear.

sub the linear into the quadratic

x^2 - 3x - 12 = y = 2x - 16

x^2 - 3x - 12 = 2x - 16

x^2 - 5x + 4 = 0

(x-1)(x-4) = 0

x = 1 or x = 4

sub into the easier of the two, namely into y = 2x - 16

if x = 1 then y = 2(1) - 16 = -14

if x = 4 then y = 2(4) - 16 = -8

so two points of intersection, (1,-14) and (4,-8)

We were lucky that the quadratic factored, if not you will have to use the quadratic equation formula.

for the second, complete the square

y=x^2-3x-12

y = x^2 - 3x + 9/4 - 9/4 - 12

= (x - 3/2)^2 - 57/4

so the min value of the function is -57/4 and it occurs when x = 3/2

or

the vertex is (3/2,-57/4)

- Mathematics -
**MathMate**, Thursday, June 25, 2009 at 11:11amHere a link to the graph of this problem:

http://i263.photobucket.com/albums/ii157/mathmate/raquel.png

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