x f x - 1 10 0 14 1 20 For the quadratic function f ,

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

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

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

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