Question

(a) Find the slope of the tangent line to the parabola \(y = {x^3} + 1\) at the point \(\left( {1,2} \right)\)

(i) using Definition 1 (ii) using Equation 2

(b) Find an equation of the tangent line in part (a).

(c) Graph the parabola and the tangent line. As a check on your work, zoom in toward the point \(\left( {1,2} \right)\) until the parabola and the tangent line are indistinguishable.

Short answer

a. The slope of the tangent line is 3.

b. The equation for tangent line is \(y = 3x - 1\)

c. The graph for curve and tangent line is:

Step-by-step solution

Slope of the Function

The slope of any function can be defined as the valueof the derivative of that function for any fixed point through which the tangent of that curve has been drwan.

(a) Step 2: Find the slope of the given function using definition 1 (i)

The function is\(y = {x^3} + 1\).

Now, using definition 1, for point (1, 2), we have:

\(\begin{aligned}m &= \mathop {\lim }\limits_{x \to 1} \frac{{f\left( x \right) - f\left( 1 \right)}}{{x - 1}}\\ &= \mathop {\lim }\limits_{x \to 1} \frac{{\left( {{x^3} + 1} \right) - \left( {1 + 1} \right)}}{{x - 1}}\\ &= \mathop {\lim }\limits_{x \to 1} \frac{{{x^3} - 1}}{{x - 1}}\\ &= \mathop {\lim }\limits_{x \to 1} \frac{{\left( {x - 1} \right)\left( {{x^2} + x + 1} \right)}}{{x - 1}}\\ &= \mathop {\lim }\limits_{x \to 1} \left( {{x^2} + x + 1} \right)\\ &= 3\end{aligned}\)

Hence, the slope of the function using definition1 is 3.

Find the slope of the given function using equation 2 (ii)

The function is\(y = {x^3} + 1\).

Now, using equation 2, for point (1,2), we have:

\(\begin{aligned}m &= \mathop {\lim }\limits_{h \to 0} \frac{{f\left( {a + h} \right) - f\left( a \right)}}{h}\\ &= \mathop {\lim }\limits_{h \to 0} \frac{{\left( {{{\left( {1 + h} \right)}^3} + 1} \right) - \left( 2 \right)}}{h}\\ &= \mathop {\lim }\limits_{h \to 0} \frac{{\left( {{h^3} + 3{h^2} + 3h - 2} \right) - \left( { - 2} \right)}}{h}\\ &= \mathop {\lim }\limits_{h \to 0} \frac{{h\left( {{h^2} + 3h + 3} \right)}}{h}\\ &= \mathop {\lim }\limits_{h \to 0} \left( {{h^2} + 3h + 3} \right)\\ &= 3\end{aligned}\)

Hence, the slope of the function using equation 2 is 3.

(b) Step 4: Find the equation of the tangent line of the function

Since the function is \(y = {x^3} + 1\)and the slope is 3, the tangent line equation is given by:

\(\begin{aligned}\left( {y - 2} \right) &= \left( {\frac{{dy}}{{dx}}\left| {_{\left( {1.2} \right)}} \right.} \right)\left( {x - 1} \right)\\y - 2 &= 3\left( {x - 1} \right)\\y - 2 &= 3x - 3\\y &= 3x - 1\end{aligned}\)

Hence, the tangent line equation is: \(y = 3x - 1\).

(c) Step 5: Sketch the graph for parabola and tangent line

Now, for the given function\(y = {x^3} + 1\), the graph for the curve and the tangent line can be drawn as: