Analyze a Quadratic Function in Vertex Form

Example: Given the quadratic function in vertex form, state the domain, range, vertex, x-intercepts, y-intercept, the orientation (opens up or opens down), and the axis of symmetry. Finally graph the function.

$f(x)=-{1/4}(x-1)^2+5$

Solution:

Find the vertex.

When the quadratic function is written in standard form you can identify the vertex as (h,k).

f(x)=a(x-h)^2+k

$f(x)=-{1/4}(x-1)^2+5$

The vertex of the quadratic function is (1,5) .

Find the orientation.

The leading coefficient of the quadratic function is negative so the parabola opens down.

Find the y-intercept.

To find a y-intercept let x=0.

Start with the original function.	$f(x)=-{1/4}(x-1)^2+5$
To find a y-intercept let x=0.	$f(0)=-{1/4}(0-1)^2+5$
Simplify using order of operations. First complete operations inside parenthesis.	$f(0)=-{1/4}(0-1)^2+5$ $f(0)=-{1/4}(-1)^2+5$
Apply the exponent.	$f(0)=-{1/4}(-1)^2+5$
Multiply.
Get a common denominator and combine fractions

The y-intercept is (0,19/4) .

Find the x-intercept.

Start with the original function.	$f(x)=-{1/4}(x-1)^2+5$
To find an x-intercept let y=0 or f(x)=0.	$0=-{1/4}(x-1)^2+5$
Isolate the square by subtracting 5 on both sides. Then simplify.	$0=-{1/4}(x-1)^2+5$ $0-5=-{1/4}(x-1)^2+5-5$ $-5=-{1/4}(x-1)^2$
Isolate the square by multiplying by -4 on both sides. Then simplify.	$-5=-{1/4}(x-1)^2$ $-5(-4)=(-4)-{1/4}(x-1)^2$
Get rid of the square by square rooting both sides.	$sqrt{20}=sqrt{(x-1)^2}$
Simplify the radical by finding a perfect square factor of 20.
Isolate x by adding 1 to both sides. Then simplify.
Rewrite as two separate answers.	or