## Algebra 2 (1st Edition)

Published by McDougal Littell

# Chapter 4 Quadratic Functions and Factoring - 4.1 Quadratic Functions in Standard Form - 4.1 Exercises - Skill Practice - Page 241: 35

#### Answer

The function has a minimum, and the value of that minimum is $y=7$.

#### Work Step by Step

The $A$-value is positive, so the function will open upwards. Therefore, the function will have a minimum. The $x$-value of the minimum's coordinate is not really useful in the answer, but we still need to find it to calculate the $y$-value of the minimum. The $x$-value of the coordinate of the function's minimum is $x=-\frac{b}{2a}$, where the function is in the form $x=Ax^{2}+Bx+C$. Thus, the value of $B$ is $8$, and the value of $A$ is $2$. Therefore: $x=-\frac{8}{2\times2}=-\frac{8}{4}=-4$ Now we can plug that $x$-value into the function, and evaluate for $y$. $y=2\times(-4)^{2}+8\times(-4)+7=32+(-32)+7=7$ Therefore, the $y$-value is $7$.

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