Follow the same procedure as Raymond, Hannah, Aidan, and Sarah did.
B
b
Follow the same procedure as Raymond, Hannah, Aidan, and Sarah did.
A
a
Graphing Form:f(x)=4(x-1.5)^2-3
Axis of Symmetry: x=1.5
Vertex: (1.5,-3)
B
b
Graphing Form:g(x)=2(x+3.5)^2-20.5
Axis of Symmetry: x=-3.5
Vertex: (-3.5,-20.5)
Practice makes perfect
a
The first step is to factor the first two terms so that the x^2-term has a coefficient of 1. In this case, that means factoring out 4.
f(x)=4x^2-12x+6 = 4(x^2-3x)+6
The coefficient in front x is -3. Half of that is -1.5 so the perfect square is (x-1.5)^2. Since we factored out 4 we should now, according to Sarah, subtract 4(-1.5)^2 to ensure that the equation is equivalent, giving us the following expression.
f(x)=4(x-1.5)^2-4(-1.5)^2+6
Let's simplify.
Now we have the function in graphing form, also known as vertex form.
f(x)=4(x-1.5)^2-3
This means that the axis of symmetry is x=1.5 and the vertex is (1.5,- 3).
b
We do the same thing as in the previous exercise, and start by factoring the first two terms.
g(x)=2x^2+14x+4= 2(x^2+7x)+4
The x-coefficient is 7, and half of that is 3.5. This means that the perfect square is (x+3.5)^2, but since we factored out 2 we need to subtract 2*3.5^2.
g(x)=2(x^2+7x)+4=2(x+3.5)^2-2* 3.5^2+4
Let's rewrite.
