identify axis of symmetry and the vertexy=x^2-4x-32
y=x^2-4x-32 is in the form y=ax^2 + bx + c. The symmetry of that form would be a vertical line with the equation
Determine first the value of 'a' and 'b'.
a=1 and b=-4. Given a=1 and b=-4 plug-in in the equation:
x=4/2 Simplifying makes
x=2 Therefore, the axis of symmetry is x=2.
To solve for the vertex, transform the equation into the standard form of parabola: y=(x-h)^2 + k
where h is the x-coordinate and k is the y-coordinate of the vertex. The vertex is (h,k). Take note of the signs of h and k in the equation. Notice that h is positive in (h,k) even it is preceeded by - in (x-h)^2. So if given is (x+h)^2, then you'll have (-h,k).
To start check x^2-4x-32 if it is a perfect square. It is a perfect square if you apply `((-b)/(2a))^2`
it will equate to the third term or 'c' term (in your case it is -32).
Is `(-(-4))/(2(1)) = -32 ?`
-4/2 is equal to 2 not -32, therefore x^2 - 4x - 32 is not a perfect square. So you'll have to make it a perfect square by doing this:
move -32 on the other side of the equation. (Moving to other side will change the sign to the opposite).
y + 32 = x^2 - 4x
Then use the formula ((-b)/(2a))^2 to solve for the term to make it a perfect square trinomial, so:
(-(-4)/(2*1))^2= 4 -----> add this to both sides of the equation.
y + 32 + 4 = x^2 - 4x + 4 Simplify and factor the right side of the equation.
y + 36 = (x-2)^2 Then make it in the standard form of a parabola: y = (x-2)^2 - 36
The sign of 36 became negative since moving it from left to right will change the sign. From that, h=2 and y = -36. Thus, the vertex is (2,-36). Be careful with the signs.
Having the vertex, you can verify the axis of symmetry. The axis of symmetry always passes through the vertex of the parabola and the x-coordinate is the equation of the axis of symmetry.