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Advanced Math / Nonlinear functions Difficulty: Medium

$x$ $x$	$f (x)$ $f (x)$
$- 1$ $negative 1$	$10$ $10$
$0$ $0$	$14$ $14$
$1$ $1$	$20$ $20$

For the quadratic function $f$ , the table shows three values of $x$ and their corresponding values of $f (x)$ . Which equation defines $f$ ?

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Explanation

Choice D is correct. The equation of a quadratic function can be written in the form $f (x) = a {(x - h)}^{2} + k$ , where $a$ , $h$ , and $k$ are constants. It’s given in the table that when $x = - 1$ , the corresponding value of $f (x)$ is $10$ . Substituting $negative 1$ for $x$ and $10$ for $f (x)$ in the equation $f (x) = a {(x - h)}^{2} + k$ gives $10 = a {(- 1 - h)}^{2} + k$ , which is equivalent to $10 = a (1 + 2 h + h^{2}) + k$ , or $10 = a + 2 a h + a h^{2} + k$ . It’s given in the table that when $x equals 0$ , the corresponding value of $f (x)$ is $14$ . Substituting $0$ for $x$ and $14$ for $f (x)$ in the equation $f (x) = a {(x - h)}^{2} + k$ gives $14 = a {(0 - h)}^{2} + k$ , or $14 = a h^{2} + k$ . It’s given in the table that when $x equals 1$ , the corresponding value of $f (x)$ is $20$ . Substituting $1$ for $x$ and $20$ for $f (x)$ in the equation $f (x) = a {(x - h)}^{2} + k$ gives $20 = a {(1 - h)}^{2} + k$ , which is equivalent to $20 = a (1 - 2 h + h^{2}) + k$ , or $20 = a - 2 a h + a h^{2} + k$ . Adding $20 = a - 2 a h + a h^{2} + k$ to the equation $10 = a + 2 a h + a h^{2} + k$ gives $30 = 2 a + 2 a h^{2} + 2 k$ . Dividing both sides of this equation by $2$ gives $15 = a + a h^{2} + k$ . Since $14 = a h^{2} + k$ , substituting $14$ for $a h^{2} + k$ into the equation $15 = a + a h^{2} + k$ gives $15 = a + 14$ . Subtracting $14$ from both sides of this equation gives $a equals 1$ . Substituting $1$ for $a$ in the equations $14 = a h^{2} + k$ and $20 = a h^{2} - 2 a h + a + k$ gives $14 = h^{2} + k$ and $20 = 1 - 2 h + h^{2} + k$ , respectively. Since $14 = h^{2} + k$ , substituting $14$ for $h^{2} + k$ in the equation $20 = 1 - 2 h + h^{2} + k$ gives $20 = 1 - 2 h + 14$ , or $20 = 15 - 2 h$ . Subtracting $15$ from both sides of this equation gives $5 = - 2 h$ . Dividing both sides of this equation by $negative 2$ gives $- \frac{5}{2} = h$ . Substituting $- \frac{5}{2}$ for $h$ into the equation $14 = h^{2} + k$ gives $14 = {(- \frac{5}{2})}^{2} + k$ , or $14 = \frac{25}{4} + k$ . Subtracting $StartFraction 25 Over 4 EndFraction$ from both sides of this equation gives $\frac{31}{4} = k$ . Substituting $1$ for $a$ , $- \frac{5}{2}$ for $h$ , and $StartFraction 31 Over 4 EndFraction$ for $k$ in the equation $f (x) = a {(x - h)}^{2} + k$ gives $f (x) = {(x + \frac{5}{2})}^{2} + \frac{31}{4}$ , which is equivalent to $f (x) = x^{2} + 5 x + \frac{25}{4} + \frac{31}{4}$ , or $f (x) = x^{2} + 5 x + 14$ . Therefore, $f (x) = x^{2} + 5 x + 14$ defines $f$ .

Choice A is incorrect. If $f (x) = 3 x^{2} + 3 x + 14$ , then when $x = - 1$ , the corresponding value of $f (x)$ is $14$ , not $10$ .

Choice B is incorrect. If $f (x) = 5 x^{2} + x + 14$ , then when $x = - 1$ , the corresponding value of $f (x)$ is $18$ , not $10$ .

Choice C is incorrect. If $f (x) = 9 x^{2} - x + 14$ , then when $x = - 1$ , the corresponding value of $f (x)$ is $24$ , not $10$ , and when $x equals 1$ , the corresponding value of $f (x)$ is $22$ , not $20$ .