This week, we learned how to find an equation of a function using the vertex and one point on the line. We then were about to put the equation into standard form.

Example 1:

Vertex= (3,-4)

Point= (4,1)

y=a(x-3)^2-4

*note: the vertex is put into the equation  The x value, 3 is made negative and put into the equation with the 4 stays negative.

Next, input the point we were given into the x and y spots in the equation, then solve for a.

(1)=a(4-3)^2-4

1=a-4

5=a

Then take your a value and complete the equation.

y=5(x-3)^2-4

Then to put it into general form, we must expand the equation.

y=5(x-3)(x-3)-4

y=(5x-15)(x-3)-4

y=5x^2-15x-15x+45-4

y=5x^2-30x+41

Example 2:

Vertex=(7,-6)

Point= (9,-4)

y=a(x-7)^2-6

-4=a(9-7)^2-6

-4=a4-6

2=a4

1/2=a

y=1/2(x-7)^2-6

=1/2(x-7)(x-7)-6

=(1/2x-7/2)(x/1-7/1)-6

=1/14x^2-7/2x-7/2x+49/2-6

=1/14x^2-14/2x-37/2

y=1/14x^2-7x-37/2