Solution
The given function is:
$$ f(x) = 8x^2 + 10x - 3 $$
(a) Write $f(x)$ in the form $A(x + B)^2 + C$
We complete the square to rewrite $f(x)$ in the desired form.
1. Factor $8$ from the quadratic and linear terms:
$$ f(x) = 8\left(x^2 + \frac{10}{8}x\right) - 3 = 8\left(x^2 + \frac{5}{4}x\right) - 3 $$
2. Complete the square inside the parentheses:
$$ x^2 + \frac{5}{4}x = \left(x + \frac{5}{8}\right)^2 - \left(\frac{5}{8}\right)^2 $$
$$ x^2 + \frac{5}{4}x = \left(x + \frac{5}{8}\right)^2 - \frac{25}{64} $$
3. Substitute back into $f(x)$:
$$ f(x) = 8\left[ \left(x + \frac{5}{8}\right)^2 - \frac{25}{64} \right] - 3 $$
$$ f(x) = 8\left(x + \frac{5}{8}\right)^2 - 8 \cdot \frac{25}{64} - 3 $$
$$ f(x) = 8\left(x + \frac{5}{8}\right)^2 - \frac{200}{64} - 3 $$
$$ f(x) = 8\left(x + \frac{5}{8}\right)^2 - \frac{25}{8} - 3 $$
$$ f(x) = 8\left(x + \frac{5}{8}\right)^2 - \frac{49}{8} $$
Thus, $f(x)$ in the form $A(x + B)^2 + C$ is:
$$ A = 8, \quad B = \frac{5}{8}, \quad C = -\frac{49}{8} $$
(b) Find the minimum value of $f(x)$
(i) The value of $x$ for which $f(x)$ has a minimum
The minimum value of a quadratic function occurs at $x = -B$. For $f(x) = A(x + B)^2 + C$:
$$ x = -B = -\frac{5}{8} $$
(ii) The minimum value of $f(x)$
Substitute $x = -\frac{5}{8}$ into the rewritten form:
$$ f(x) = 8\left(x + \frac{5}{8}\right)^2 - \frac{49}{8} $$
At $x = -\frac{5}{8}$:
$$ f(x) = 8(0)^2 - \frac{49}{8} = -\frac{49}{8} $$
Thus, the minimum value of $f(x)$ is:
$$ \boxed{-\frac{49}{8}} $$
(c) Find the $x$-coordinates of the points where $C$ crosses the $x$-axis
At the $x$-axis, $f(x) = 0$:
$$ 8x^2 + 10x - 3 = 0 $$
Use the quadratic formula:
$$ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}, \quad a = 8,\; b = 10,\; c = -3 $$
$$ x = \frac{-10 \pm \sqrt{10^2 - 4(8)(-3)}}{2(8)} $$
$$ x = \frac{-10 \pm \sqrt{100 + 96}}{16} $$
$$ x = \frac{-10 \pm \sqrt{196}}{16} $$
$$ x = \frac{-10 \pm 14}{16} $$
Solve for the two roots:
$$ x = \frac{-10 + 14}{16} = \frac{4}{16} = \frac{1}{4} $$
$$ x = \frac{-10 - 14}{16} = \frac{-24}{16} = -\frac{3}{2} $$
The $x$-coordinates of the points where $C$ crosses the $x$-axis are:
$$ \boxed{ x = \frac{1}{4} \text{ and } x = -\frac{3}{2} } $$
(d) Find the points of intersection of $C$ and $l$
The curve $C$ has equation $y = f(x)$ and the line $l$ has equation $y = 2x + 13$. Set the two equations equal:
$$ 8x^2 + 10x - 3 = 2x + 13 $$
Rearrange into standard quadratic form:
$$ 8x^2 + 10x - 3 - 2x - 13 = 0 $$
$$ 8x^2 + 8x - 16 = 0 $$
Simplify:
$$ x^2 + x - 2 = 0 $$
Factorize:
$$ x^2 + x - 2 = (x + 2)(x - 1) = 0 $$
Solve for $x$:
$$ x = -2 \quad \text{or} \quad x = 1 $$
Find the corresponding $y$-coordinates using $y = 2x + 13$:
$$ \text{For } x = -2: \quad y = 2(-2) + 13 = -4 + 13 = 9 $$
$$ \text{For } x = 1: \quad y = 2(1) + 13 = 2 + 13 = 15 $$
The points of intersection are:
$$ \boxed{ (-2, 9) \text{ and } (1, 15) } $$
Sketch of the Curve $C$ and the Line $l$
The curve $C$ has equation $y = 8x^2 + 10x - 3$, and the line $l$ has equation $y = 2x + 13$. Using the results:
- From part (b), the minimum value of $f(x)$ is $-\frac{49}{8}$ at $x = -\frac{5}{8}$.
- From part (c), $C$ crosses the $x$-axis at $x = \frac{1}{4}$ and $x = -\frac{3}{2}$.
- From part (d), $C$ and $l$ intersect at points $(-2, 9)$ and $(1, 15)$.
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