Find the zeros and the maximum or minimum value of the function

A function is given. Plot the zeros and the maximum or minimum value of the function.

​f(x) = (2x- 2)(x - 3)

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To find the zeros of a function (where the graph intercepts the x-axis), set the value of y equal to zero, and solve for x.

f(x) is the y-value.

f(x) = (2x - 2)(x - 3)
0 = (2x - 2)(x - 3) {set the value of y equal to zero}
2x - 2 = 0 or x - 3 = 0 {if two or more factors multiply together to equal zero, then either one could be equal to zero}
2x = 2 or x = 3 {set each factor equal to zero}
x = 1 or x = 3 {solved for x}

The zeros (x-intercepts) are 1 and 3. To plot, put a point at 1 on the x-axis and a point at 3 on the x-axis.

If the coefficient of x² is positive, the parabola opens upward, meaning there is a minimum value.
If the coefficient of x² is negative, the parabola opens downward, meaning there is a maximum value.

To find the maximum or minimum (the y-coordinate of the vertex), get the equation in standard form.

Standard form of a quadratic function is ax² + bx + c.

f(x) = (2x - 2)(x - 3)
f(x) = 2x² - 6x - 2x + 6 {used foil method or distributive property}
f(x) = 2x² - 8x + 6 {combined like terms}

The x-coordinate of the vertex is x = -b/2a

a is the coefficient of x² and b is the coefficient of x.

a = 2 and b = -8

-b
—- = x {the x-coordinate of the vertex}
2a

-(-8)
——— = x {substituted 2 for a and -8 for b}
2(2)

8
— = x {simplified}
4

x = 2 {divided}

To find the y-coordinate of the vertex. Substitute the x-coordinate of the vertex back into the equation.

y = 2x² - 8x + 6
y = 2(2)² - 8(2) + 6 {substituted 2 for x}
y = 2(4) - 16 + 6 {evaluated exponent and multiplied}
y = 8 - 16 + 6 {multiplied 2 by 4}
y = -2 {added and subtracted}

The vertex has coordinates of (2,-2)

Since the coefficient of x² is positive, there is a minimum value. That minimum value is the y-coordinate of the vertex.

The minimum value is -2. The coordinates of the minimum value are (2,-2). To plot, put a point at (2,-2), as shown below.

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