How do you find the vertex of the graph of a quadratic function written in intercept form?

Intercept form is also known as factored form: y=(x-p)(x-q) where p,q are the x-intercepts.

One way to find the vertex is to rewrite in standard form: the vertex is located at `((-b)/(2a),f((-b)/(2a))) ` where y=f(x).

Slightly more efficient is to do the following:

Recognize that the graph is symmetric about the axis of symmetry. The equation for the axis of symmetry is `x=-b/(2a) ` . Since the graph is symmetric, the zeros are the same distance away from the axis of symmetry. More to the point, the axis is the arithmetic mean of the two zeros. So the equation for the axis of symmetry can also be written as `x=(p+q)/2 ` where p,q are the x-intercepts.

For example, y=(x-2)(x+4). The zeros are x=2,x=-4. The average of the zeros is (2-4)/2=-1, so the axis of symmetry is x=-1.

Once you have the equation for the axis of symmetry, you have the x coordinate of the vertex. For the example given, the x coordinate of the vertex is -1. Substitute this into the given equation to find the y coordinate. Here y=(-1-2)(-1+4)=(-3)(3)=-9 so the vertex is at (-1,-9).

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To find the vertex, the x coordinate is the average of the zeros; substitute this value for x to get the y coordinate.

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The graph of y=(x-2)(x+4):