Graph y=x^2-4

Problem

y=x2−4

Solution

1. Identify the type of function. This is a quadratic function in the form y=a*x2+c which represents a parabola opening upwards because a=1 is positive.

2. Determine the vertex. Since there is no x term (b=0, the xcoordinate of the vertex is x=0 Substituting x=0 into the equation gives y=−4 so the vertex is (0,−4)

3. Find the xintercepts by setting y=0

0=x2−4

x2=4

x=±2

The xintercepts are (2,0) and (−2,0)

4. Find the yintercept by setting x=0

y=0−4

y=−4

The yintercept is (0,−4) which is also the vertex.

5. Plot additional points to define the shape. For x=1 and x=−1

y=(1)2−4=−3

y=(−1)2−4=−3

The points are (1,−3) and (−1,−3)

6. Sketch the curve. Draw a smooth U-shaped parabola passing through the points (−2,0) (−1,−3) (0,−4) (1,−3) and (2,0)

Final Answer

y=x2−4

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