A function is given. Plot the zeros and the maximum or minimum value of the function.
f(x) = (2x 2)(x  3)
To find the zeros of a function (where the graph intercepts the xaxis), set the value of y equal to zero, and solve for x.
f(x) is the yvalue. 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 (xintercepts) are 1 and 3. To plot, put a point at 1 on the xaxis and a point at 3 on the xaxis. 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 ycoordinate 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 xcoordinate 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 xcoordinate 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 ycoordinate of the vertex. Substitute the xcoordinate 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 ycoordinate 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. Ask Algebra House Comments are closed.

Latest Videos
Algebra 1 State Test Practice Archives
November 2023
