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In the $x y$-plane, the graph of the function $f(x)=x^{2}+5 x+4$ has two $x$-intercepts. What is the distance between the $x$-intercepts?

SAT · Math · previous-year question

  1. A.1
  2. B.2
  3. C.3correct
  4. D.4

Answer

C. 3

Explanation

Choice $\mathbf{C}$ is correct. The $x$-intercepts of the graph of $f(x)=x^{2}+5 x+4$ are the points $(x, f(x))$ on the graph where $f(x)=0$. Substituting 0 for $f(x)$ in the function equation yields $0=x^{2}+5 x+4$. Factoring the right-hand side of $0=x^{2}+5 x+4$ yields $0=(x+4)(x+1)$. If $0=(x+4)(x+1)$, then $0=x+4$ or $0=x+1$. Solving both of these equations for $x$ yields $x=-4$ and $x=-1$. Therefore, the $x$-intercepts of the graph of $f(x)=x^{2}+5 x+4$ are $(-4,0)$ and $(-1,0)$. Since both points lie on the $x$-axis, the distance between $(-4,0)$ and $(-1,0)$ is equivalent to the number

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