Find the minimum value of the curve 12x² + 24x + 13.

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 $\boxed{-7}$.