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
f(0)=-{1/4}(1)+5
Multiply.
f(0)=-{1/4}(1)+5
f(0)=-{1/4}+5
Get a common denominator and combine fractions
f(0)=-{1/4}+5
f(0)=-{1/4}+20/4
f(0)=19/4

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
20=(x-1)^2
Get rid of the square by square rooting both sides.
20=(x-1)^2
sqrt{20}=sqrt{(x-1)^2}
pm sqrt{20}=x-1
Simplify the radical by finding a perfect square factor of 20.
pm sqrt{20}=x-1
pm sqrt{4*5}=x-1
pm 2 sqrt{5}=x-1
Isolate x by adding 1 to both sides.  Then simplify.
pm 2 sqrt{5}=x-1
1 pm 2 sqrt{5}=x-1+1
1  pm   2 sqrt{5}=x
Rewrite as two separate answers.
1  pm   2 sqrt{5}=x
1 + 2 sqrt{5}=x or 1 - 2 sqrt{5}=x

The x-intercepts are (1 + 2 sqrt{5},0) and (1 - 2 sqrt{5},0)

Find the axis of symmetry.

The axis of symmetry is a vertical line that passes through the vertex.  Since the vertex is (1,5) the axis of symmetry is x=1.

Graph the function.

Plot the intercepts and the vertex.

Since the vertex is the highest point we can draw the parabola using the peak at the vertex.

Find the domain and range.

The domain is (- infty, infty).

The range is (  - infty, 5 {}{]}.

Here is a video example analyzing a quadratic function in vertex form.