LEAP 2025 Algebra I Additional & Supporting Content. Practice it free.

Set $f(x) = 0$: $2(x - 6)(x - 1)(x + 10) = 0$. The constant $2 \neq 0$, so the zeros are $x = 6$, $x = 1$, and $x = -10$. Among the choices, only $-10$ is an x-coordinate of an x-intercept.

The equation is in vertex form $f(x) = a(x - h)^2 + k$ with $a = 1 > 0$, so the graph opens upward and the minimum occurs at the vertex. Here $h = 5$, so the vertex is at $x = 5$. Therefore, $f(x)$ reaches its minimum when $x = 5$.

The parabola opens upward, so the smallest $y$ is at the vertex. For $y = ax^2 + bx + c$, the $x$-value at the vertex is $x = -\dfrac{b}{2a}$. Here $a = 2$ and $b = 12$, so $x = -\dfrac{12}{2(2)} = -\dfrac{12}{4} = -3$. Therefore, $y$ reaches its minimum when $x = -3$.

Here $a = 1$, $b = 0$, and $c = -9$. Substituting into $b^{2} - 4ac$ gives $0^{2} - 4(1)(-9) = 36\textsf{,}$ so there are $2$ distinct real roots.

Here $a = 1$, $b = -5$, and $c = 6$. Substituting into the quadratic formula gives $x = \dfrac{-(-5) \pm \sqrt{(-5)^{2} - 4(1)(6)}}{2(1)} = \dfrac{5 \pm 1}{2}$, so $x = 2$ and $x = 3$.

The equation is already $(x - 4)^{2} = -16\textsf{.}$ A real number squared cannot be negative, so there are $0$ distinct real solutions.

The quadratic factors as $(2x + 7)(x - 5) = 0\textsf{.}$ By the zero-product property, $2x + 7 = 0$ or $x - 5 = 0\textsf{,}$ so $x = -\dfrac{7}{2}$ or $x = 5\textsf{.}$ The positive solution is $5\textsf{.}$

The $y$-intercept is the value of $y$ when $x = 0\textsf{.}$ Substituting gives $3y = -12\textsf{,}$ so $y = -4\textsf{.}$