Chapter 1.2, Problem 13GP

Solution

To write: A quadratic function in standard form.

The standard form of the function $y=-3{\left(x+5\right)}^2-1$ is $y=-3x^2-30x-76$ .

Given data: The quadratic function form is $y=-3{\left(x+5\right)}^2-1$ .

Method/Formula used:

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

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

If $c=a,d=b$ then, $P=(a+b)(a+b)=a^2+2a\cdot b+b^2$

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 $y=-3{\left(x+5\right)}^2-1$ can be written in standard form as,

  $\begin{array}{c}y=-3\left[{\left(x+5\right)}^2\right]-1\leftarrow \text{Original}\text{form}\\ =-3\left[x\cdot x+2x\cdot 5+5^2\right]-1\leftarrow \text{Use square}\text{formula}\\ =-3\left[x^2+10x+25\right]-1\leftarrow \text{Simplify}\\ =-3\left[x^2+10x+25\right]-1\leftarrow \text{Use}\text{distributive}\text{property}\\ =\left[\left(-3\right)\cdot x^2+\left(-3\right)\cdot \left(10x\right)+\left(-3\right)\cdot \left(25\right)\right]-1\\ =-3x^2-30x-75-1\\ =-3x^2-30x-76\underset{¯}{\text{Standard}\text{form}}\end{array}$

Thus, the standard form of the function $y=-3{\left(x+5\right)}^2-1$ is $y=-3x^2-30x-76$ .