Algebra 2 Common Core

Published by Prentice Hall
ISBN 10: 0133186024
ISBN 13: 978-0-13318-602-4

Chapter 6 - Radical Functions and Rational Exponents - Cumulative Standards Review - Multiple Choice - Page 430: 21

Answer

The $x$-value of the $x$-intercept is $-2$.

Work Step by Step

The definition of the $x$-intercept is the value of $x$ when $y$ equals $0$. To find the $x$ intercept of the graph given, we set $y$ or $f(x)$ equal to $0$: $0 = x^2 + 4x + 4$ To solve for $x$, we try to factor first: To factor a quadratic polynomial in the form $x^2 + bx + c$, we look at factors of $c$ such that, when added together, equal $b$. For the trinomial $x^2 + 4x + 4$, $c=4$, so look for factors of $4$ that when added together will equal $b$ or $4$. Both factors must be positive. Here are the possibilities: $4=(4)(1)$ $4+1 = 5$ $4=(2)(2)$ $2+2 = 4$ The second pair, $2$ and $2$, is what we are looking for. Hence, the factored of the trinomial is $(x + 2)(x + 2)$. Thus, the equation above is equivalent to: $$(x+2)(x+2)=0$$ According to the zero product property, if the product of two factors $a$ and $b$ equals zero, then either $a$ is zero, $b$ is zero, or both equal zero. Therefore, we can set each factor equal to zero; however, since both factors are the same, we solve for $x$ only once: The first factor: $x + 2 = 0$ Subtract $2$ from each side to solve for $x$: $x = -2$ This is the $x$-value of the $x$-intercept.
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