identify axis of symmetry and the vertexy=x^2-4x-32

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mariloucortez | High School Teacher | (Level 3) Adjunct Educator

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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.