The diagram opposite shows graphs of the quadratic function $f(x) = x^2 + 3x - 1$, $x \, ext{in} \, ext{R}$ and the line $l_1$. The line $l_1$ passes through the point $(2, 0)$ and is a tangent to the curve at the point $(-1, -3)$. (a) Find the slope of $l_1$, using a slope formula. (b) (i) Find $f'(x)$, the derivative of $f(x)$. (ii) Verify your answer to (a) above by finding the value of $f'(x)$ at $x = -1$. (c) The line $l_2$ is perpendicular to $l_1$ and is also a tangent to the curve $f(x)$. Find the co-ordinates of the point at which $l_2$ touches the curve.

Leaving Cert Mathematics Calculus - Differentiation: The diagram opposite shows graph

The diagram opposite shows graphs of the quadratic function $f(x) = x^2 + 3x - 1$, $x \, ext{in} \, ext{R}$ and the line $l_1$ - Leaving Cert Mathematics - Question 1 - 2013

Question 1

$The-diagram-opposite-shows-graphs-of-the-quadratic-function-$f(x)-=-x^2-+-3x---1$,-$x-\,--ext{in}-\,--ext{R}$-and-the-line-$l_1$-Leaving Cert Mathematics-Question 1-2013.png$

The diagram opposite shows graphs of the quadratic function $f(x) = x^2 + 3x - 1$, $x \, ext{in} \, ext{R}$ and the line $l_1$. The line $l_1$ passes through the p... show full transcript

Worked Solution & Example Answer:The diagram opposite shows graphs of the quadratic function $f(x) = x^2 + 3x - 1$, $x \, ext{in} \, ext{R}$ and the line $l_1$ - Leaving Cert Mathematics - Question 1 - 2013

Step 1

Find the slope of $l_1$, using a slope formula.

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Answer

To calculate the slope of the line $l_1$ , we use the formula:

$ext{slope} = \frac{y_2 - y_1}{x_2 - x_1}$

Substituting the points $(-1, -3)$ and $(2, 0)$ :

$ext{slope} = \frac{0 - (-3)}{2 - (-1)} = \frac{3}{3} = 1$

Therefore, the slope of $l_1$ is 1.

Step 2

Find $f'(x)$, the derivative of $f(x)$.

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Answer

To find the derivative of the function $f(x) = x^2 + 3x - 1$ , we differentiate with respect to $x$ :

$f'(x) = \frac{d}{dx}(x^2 + 3x - 1) = 2x + 3$

Step 3

Verify your answer to (a) above by finding the value of $f'(x)$ at $x = -1$.

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Answer

We substitute $x = -1$ into the derivative:

$f'(-1) = 2(-1) + 3 = -2 + 3 = 1$

This confirms that the slope of the tangent line $l_1$ at $x = -1$ is indeed 1.

Step 4

The line $l_2$ is perpendicular to $l_1$ and is also a tangent to the curve $f(x)$. Find the co-ordinates of the point at which $l_2$ touches the curve.

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Answer

Since the slope of $l_1$ is 1, the slope of the perpendicular line $l_2$ will be:

$m_{l_2} = -\frac{1}{1} = -1$

The equation of the line $l_2$ can be found using point-slope form. Since $l_1$ touches at $(-1, -3)$ :

$y - (-3) = -1(x - (-1))$ Simplifying gives:

ightarrow y = -x - 4$$ Now, we need the point of tangency with the curve, where $y = -x - 4$ satisfies $f(x)$: $$x^2 + 3x - 1 = -x - 4$$ Rearranging results in: $$x^2 + 4x + 3 = 0$$ Factoring: $$(x + 1)(x + 3) = 0$$ Thus, $x = -1$ or $x = -3$. For $x = -3$: Substituting back to find $y$: $$y = -(-3) - 4 = -3$$ Therefore, the point of contact where $l_2$ touches the curve is $(-2, -3)$.

The diagram opposite shows graphs of the quadratic function $f(x) = x^2 + 3x - 1$, $x \, ext{in} \, ext{R}$ and the line $l_1$ - Leaving Cert Mathematics - Question 1 - 2013

Worked Solution & Example Answer:The diagram opposite shows graphs of the quadratic function $f(x) = x^2 + 3x - 1$, $x \, ext{in} \, ext{R}$ and the line $l_1$ - Leaving Cert Mathematics - Question 1 - 2013

Find the slope of $l_1$, using a slope formula.

Find $f'(x)$, the derivative of $f(x)$.

Verify your answer to (a) above by finding the value of $f'(x)$ at $x = -1$.

The line $l_2$ is perpendicular to $l_1$ and is also a tangent to the curve $f(x)$. Find the co-ordinates of the point at which $l_2$ touches the curve.

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