MTH 065 Elementary Algebra II Chapter 11 Quadratic Functions and Equations Section 11.7 More About Graphing Quadratic Functions

Forms of Quadratic Functions Standard form … f(x) = ax 2 + bx + c Graphing form … f(x) = a(x – h) 2 + k How to convert between these forms …

Graphing Form  Standard From Square the binomial, remove grouping symbols, and combine like terms … f(x) = 2(x – 4) f(x) = 2(x 2 – 8x + 16) + 5 f(x) = 2x 2 – 16x f(x) = 2x 2 – 16x + 37

Standard Form  Graphing From Factor a out of the terms w/ the variable, complete the square, factor, and add the constants … f(x) = 2x 2 – 16x + 37 f(x) = 2(x 2 – 8x) + 37 f(x) = 2(x 2 – 8x ) + 37 – 32 f(x) = 2(x – 4) 2 + 5

Locating the Vertex from the Standard From f(x) = 2x 2 – 16x + 37 f(x) = 2(x 2 – 8x) + 37 f(x) = 2(x 2 – 8x ) + 37 – 32 f(x) = 2(x – 4) To get to the 4 (the x-coordinate of the vertex) … Start with the opposite of the linear coefficient (b) … Divide by the quadratic coefficient (a) … Divide by 2 i.e.

Locating the Vertex from the Standard From f(x) = 2x 2 – 16x + 37 f(x) = 2(x 2 – 8x) + 37 f(x) = 2(x 2 – 8x ) + 37 – 32 f(x) = 2(x – 4) To get to the 5 (the y-coordinate of the vertex) … Start with the constant term (c) … Subtract a times the square of the x-coordinate … i.e.

Locating the Vertex f(x) = ax 2 + bx + c Vertex:

The y-intercept Let x = 0 f(x) = ax 2 + bx + c f(0) = a(0) 2 + b0 + c f(0) = c Therefore, the constant term is the y-intercept.

The x-intercepts Let f(x) = 0 f(x) = ax 2 + bx + c 0 = ax 2 + bx + c … solve for x (if possible) … Therefore, there may be 0, 1, or 2 x-intercepts.