NSC Mathematics Calculus: Given: $f(x) = -ax^2 + bx + 6$

Given: $f(x) = -ax^2 + bx + 6$ 4.1 The gradient of the tangent to the graph of $f$ at the point $(-1, \frac{7}{2})$ is 3 - NSC Mathematics - Question 4 - 2017 - Paper 1

Question 4

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Given: $f(x) = -ax^2 + bx + 6$ 4.1 The gradient of the tangent to the graph of $f$ at the point $(-1, \frac{7}{2})$ is 3. Show that $a = \frac{1}{2}$ and $b =... show full transcript

Worked Solution & Example Answer:Given: $f(x) = -ax^2 + bx + 6$ 4.1 The gradient of the tangent to the graph of $f$ at the point $(-1, \frac{7}{2})$ is 3 - NSC Mathematics - Question 4 - 2017 - Paper 1

Step 1

The gradient of the tangent to the graph of $f$ at the point $(-1, \frac{7}{2})$ is 3.

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Answer

To find the gradient, we need to differentiate the function:
$f'(x) = -2ax + b$ .
Evaluating at $x = -1$ :
$f'(-1) = -2a(-1) + b = 2a + b$ .
We know this equals 3. Thus, we have:
$2a + b = 3$ .

We also know that $f(-1) = \frac{7}{2}$ :
$f(-1) = -a(-1)^2 + b(-1) + 6.$
Substituting, we get:
$-a - b + 6 = \frac{7}{2}$ , which simplifies to:
$-a - b = \frac{7}{2} - 6 = -\frac{5}{2}$ .
Therefore, we have two equations:

$2a + b = 3$
$-a - b = -\frac{5}{2}$ .

To solve, we can first simplify the second equation:
$a + b = \frac{5}{2}.$
Now, we can use substitution or elimination to find $a$ and $b$ .
Substituting $b = 3 - 2a$ into the second equation yields:
$a + (3 - 2a) = \frac{5}{2}$ , leading to:
$-a + 3 = \frac{5}{2},$ resulting in:
$-a = \frac{5}{2} - 3 = -\frac{1}{2},$ thus:
$a = \frac{1}{2}.$
Substituting $a$ back into the first equation gives $b = 2$ .

Step 2

Calculate the $x$-intercepts of $f$.

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Answer

To find the $x$ -intercepts, set $f(x) = 0$ :
$-\frac{1}{2}x^2 + 2x + 6 = 0.$
Multiplying by -2 to eliminate the fraction:
$x^2 - 4x - 12 = 0.$
Factoring this gives:
$(x - 6)(x + 2) = 0.$
Thus, the $x$ -intercepts are $x = 6$ and $x = -2$ , or in coordinate form, $(6, 0)$ and $(-2, 0)$ .

Step 3

Calculate the coordinates of the turning point of $f$.

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Answer

The turning point can be found using $f'(x) = 0$ :
$-2a x + b = 0.$
Substituting $a$ and $b$ , we have:
$-x + 2 = 0,$ leading to:
$x = 2.$
Now, substituting $x = 2$ back into $f(x)$ to find the corresponding $y$ value:
$f(2) = -\frac{1}{2}(2)^2 + 2(2) + 6 = -2 + 4 + 6 = 8.$
Thus, the turning point is at $(2, 8)$ .

Step 4

Sketch the graph of $f$. Clearly indicate ALL intercepts with the axes and the turning point.

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Answer

$f$ is a downward-opening parabola with intercepts at $(6,0)$ and $(-2,0)$ , and a turning point at $(2, 8)$ . Continue the sketch by marking the axes to show these points.

Step 5

Use the graph to determine the values of $x$ for which $f(x) > 6$.

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Answer

From the graph, we see that $f(x) > 6$ in the interval $( -2, 6)$ except at $x = 2$ , where $f(x) = 8$ . Thus, $x$ values for which $f(x) > 6$ are $x ext{ in } (-2, 2) ext{ and } (2, 6)$ .

Step 6

Sketch the graph of $g(x) = -x - 1$ on the same set of axes as $f$. Clearly indicate ALL intercepts with the axes.

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Answer

The line $g(x)$ has a $y$ -intercept at $(0, -1)$ and an $x$ -intercept at $(-1, 0)$ . Mark these points on the graph along with the parabola for $f$ .

Step 7

Write down the values of $x$ for which $f(x)g(x) \leq 0$.

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Answer

To solve $f(x)g(x) \leq 0$ , we analyze the intervals where $f(x)$ and $g(x)$ have opposite signs. Points of interest are $(2, 8)$ and the intercepts. This leads us to determine $x$ values that satisfy the condition between the $y$ -axis intercepts.

Given: $f(x) = -ax^2 + bx + 6$ 4.1 The gradient of the tangent to the graph of $f$ at the point $(-1, \frac{7}{2})$ is 3 - NSC Mathematics - Question 4 - 2017 - Paper 1

Worked Solution & Example Answer:Given: $f(x) = -ax^2 + bx + 6$ 4.1 The gradient of the tangent to the graph of $f$ at the point $(-1, \frac{7}{2})$ is 3 - NSC Mathematics - Question 4 - 2017 - Paper 1

The gradient of the tangent to the graph of $f$ at the point $(-1, \frac{7}{2})$ is 3.

Calculate the $x$-intercepts of $f$.

Calculate the coordinates of the turning point of $f$.

Sketch the graph of $f$. Clearly indicate ALL intercepts with the axes and the turning point.

Use the graph to determine the values of $x$ for which $f(x) > 6$.

Sketch the graph of $g(x) = -x - 1$ on the same set of axes as $f$. Clearly indicate ALL intercepts with the axes.

Write down the values of $x$ for which $f(x)g(x) \leq 0$.

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