Chapter 1.2, Problem 12GP

Solution

To write: A quadratic function in standard form.

The standard form of the function $y=-7\left(x-6\right)\left(x+1\right)$ is $y=-7x^2+35x+42$ .

Given data: The quadratic function form is $y=-7\left(x-6\right)\left(x+1\right)$ .

Method/Formula used:

The product of two factors $P=(x-b)(x-c)$ using FOIL method is given by

  $\begin{array}{c}P=(a+b)(c+d)\\ =\left(a\cdot c+a\cdot d+b\cdot c+b\cdot d\right)\end{array}$

Distributive property over addition: $a\cdot \left(b+c\right)=a\cdot b+a\cdot c$ .

The form of a quadratic function $y\left(x\right)=A{x}^2+Bx+C$ is called the standard form.

Calculation:

The form of quadratic function $y=-7\left(x-6\right)\left(x+1\right)$ can be written in standard form as,

  $\begin{array}{c}y=-7\left[\left(x-6\right)\left(x+1\right)\right]\leftarrow \text{Original}\text{form}\\ =-7\left[x\cdot x+x\cdot 1+\left(-6\right)\cdot x+\left(-6\right)\cdot 1\right]\leftarrow \text{Use FOIL}\text{formula}\\ =-7\left[x^2+x-6x-6\right]\leftarrow \text{Simplify}\\ \text{=}-7\left[x^2+x\left(1-6\right)-6\right]\leftarrow \text{ collect terms of like powers of }x\\ =-7\left[x^2-5x-6\right]\leftarrow \text{Use}\text{distributive}\text{property}\\ =\left(-7\right)\cdot x^2+\left(-7\right)\cdot \left(-5x\right)+\left(-7\right)\cdot \left(-6\right)\\ =-7x^2+35x+42\underset{¯}{\text{Standard}\text{form}}\end{array}$

Thus, the standard form of the function $y=-7\left(x-6\right)\left(x+1\right)$ is $y=-7x^2+35x+42$ .