In this form, where a ≠0, the x-intercepts are p and q. Let's consider the intercept form of our function.
y=(x-2)(x+2) ⇔ y= 1(x- 2)(x-( - 2))
We can see that a= 1, p= 2, and q= -2. Therefore, the x-intercepts occur at ( 2,0) and ( - 2,0).
Find and Graph the Axis of Symmetry
The axis of symmetry is halfway between (p,0) and (q,0). Since we know that p=2 and q=- 2, the axis of symmetry of our parabola is halfway between (2,0) and (- 2,0).
x=p+q/2 ⇒ x=2+( - 2)/2=0/2=0
We found that the axis of symmetry is the vertical line x=0.
Find and Plot the Vertex
Since the vertex lies on the axis of symmetry, its x-coordinate is 0. To find the y-coordinate, we will substitute 0 for x in the given equation.
The y-coordinate of the vertex is - 4. Therefore, the vertex is the point (0,- 4).
Draw the Parabola
Finally, we will draw the parabola through the vertex and the x-intercepts.
We can see above that there are no restrictions on the x-variable. Furthermore, the y-variable takes values greater than or equal to - 4. We can write the domain and range of the function using this information.
Domain:& All real numbers
Range:& y ≥ - 4
