LEAP 2025 Algebra I Major Content. Practice it free.

Point-slope form is $y - y_1 = m(x - x_1)\textsf{.}$ The given point is $(x_1, y_1) = (-3, 4)\textsf{.}$

Substitute $y = 6$. From $x + y \le 10$, $x \le 4$. From $x \ge y - 2$, $x \ge 4$. The only listed value that satisfies both is $x = 4$.

Set $y = 0$: $(x - 15)(x + 7) = 0$. The zeros are $x = 15$ and $x = -7$. Of the choices, $15$ 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 $h = 9$, so the vertex is at $x = 9$. Therefore, $f(x)$ reaches its minimum when $x = 9$.

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 $a = 1$ and $b = 4$, so $x = -\dfrac{4}{2(1)} = -\dfrac{4}{2} = -2$. Therefore, $y$ is as small as possible when $x = -2$.

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

Here $a = 1$, $b = -4$, and $c = 4$. Substituting into $b^{2} - 4ac$ gives $(-4)^{2} - 4(1)(4) = 0$, so there is exactly $1$ distinct real solution.

Subtract $12$: $x^{2} = -12\textsf{.}$ A real number squared cannot be negative, so there are $0$ distinct real solutions.