10. Find the minimum value of the curve $12x^2 + 24x + 13$. Step 1: Identify the type of curve. The given curve is a quadratic function $f(x) = 12x^2 + 24x + 13$. Since the coefficient of $x^2$ is $12$ (which is positive), the parabola opens upwards, meaning it has a minimum value at its vertex. Step 2: Find the x-coordinate of the vertex. For a quadratic function $ax^2 + bx + c$, the x-coordinate of the vertex is given by the formula $x = -\frac{b}{2a}$. Here, $a=12$ and $b=24$. $$ x = -\frac{24}{2(12)} = -\frac{24}{24} = -1 $$ Step 3: Find the minimum value (y-coordinate of the vertex). Substitute the x-coordinate of the vertex ($x=-1$) back into the function $f(x)$: $$ f(-1) = 12(-1)^2 + 24(-1) + 13 $$ $$ f(-1) = 12(1) - 24 + 13 $$ $$ f(-1) = 12 - 24 + 13 $$ $$ f(-1) = -12 + 13 $$ $$ f(-1) = 1 $$ The minimum value of the curve is $1$. The minimum value of the curve is $\boxed{1}$. 11. If the roots of $ax^2 + bx + c = 0$ are $\alpha$ and $\beta$, show that $\alpha + \beta = -\frac{b}{a}$ and $\alpha\beta = \frac{c}{a}$. Step 1: Express the quadratic equation in terms of its roots. If $\alpha$ and $\beta$ are the roots of the quadratic equation $ax^2 + bx + c = 0$, then the polynomial can be factored as $a(x-\alpha)(x-\beta) = 0$. Step 2: Expand the factored form. $$ a(x-\alpha)(x-\beta) = a(x^2 - \beta x - \alpha x + \alpha\beta) $$ $$ = a(x^2 - (\alpha+\beta)x + \alpha\beta) $$ $$ = ax^2 - a(\alpha+\beta)x + a\alpha\beta $$ Step 3: Compare coefficients with the original equation. We have $ax^2 - a(\alpha+\beta)x + a\alpha\beta = 0$. Comparing this to the original equation $ax^2 + bx + c = 0$: • Comparing the coefficients of $x$: $$ -a(\alpha+\beta) = b $$ $$ \alpha+\beta = -\frac{b}{a} $$ • Comparing the constant terms: $$ a\alpha\beta = c $$ $$ \alpha\beta = \frac{c}{a} $$ Thus, we have shown that $\alpha + \beta = -\frac{b}{a}$ and $\alpha\beta = \frac{c}{a}$. 12. Find $b$ if $h(x) = 2x-1$ is a factor of $f(x) = 6x^2 + bx + 2$. Step 1: Apply the Factor Theorem. According to the Factor Theorem, if $(2x-1)$ is a factor of $f(x)$, then $f(c) = 0$, where $c$ is the root of $2x-1=0$. Step 2: Find the root of $h(x)$. Set $h(x) = 0$: $$ 2x-1 = 0 $$ $$ 2x = 1 $$ $$ x = \frac{1}{2} $$ Step 3: Substitute the root into $f(x)$ and solve for $b$. Since $x=\frac{1}{2}$ is a root, $f\left(\frac{1}{2}\right)$ must be $0$. $$ f\left(\frac{1}{2}\right) = 6\left(\frac{1}{2}\right)^2 + b\left(\frac{1}{2}\right) + 2 = 0 $$ $$ 6\left(\frac{1}{4}\right) + \frac{b}{2} + 2 = 0 $$ $$ \frac{6}{4} + \frac{b}{2} + 2 = 0 $$ $$ \frac{3}{2} + \frac{b}{2} + 2 = 0 $$ Multiply the entire equation by $2$ to eliminate denominators: $$ 2\left(\frac{3}{2}\right) + 2\left(\frac{b}{2}\right) + 2(2) = 2(0) $$ $$ 3 + b + 4 = 0 $$ $$ b + 7 = 0 $$ $$ b = -7 $$ The value of $b$ is $\boxed{-7}$.