Posted by Raquel on Thursday, June 25, 2009 at 10:35am.
in 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 + 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
the vertex is (3/2,-57/4)
Here a link to the graph of this problem:
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