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Find the equation of the tangent to the curve $y = 2x^3 + 3$ at the point where $x = -2.$ - Scottish Highers Maths - Question 2 - 2015

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Find the equation of the tangent to the curve $y = 2x^3 + 3$ at the point where $x = -2.$

Worked Solution & Example Answer:Find the equation of the tangent to the curve $y = 2x^3 + 3$ at the point where $x = -2.$ - Scottish Highers Maths - Question 2 - 2015

Step 1

Differentiate the curve

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Answer

To find the equation of the tangent, we first need to differentiate the curve y=2x3+3y = 2x^3 + 3 to find the gradient of the tangent. The derivative is given by:

dydx=6x2\frac{dy}{dx} = 6x^2

Step 2

Evaluate the derivative at $x = -2$

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Answer

Now, we substitute x=−2x = -2 into the derivative to find the slope of the tangent:

dydx∣x=−2=6(−2)2=6⋅4=24\frac{dy}{dx}|_{x=-2} = 6(-2)^2 = 6 \cdot 4 = 24

Step 3

Find the y-coordinate at $x = -2$

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Answer

Next, we need to find the corresponding yy-coordinate when x=−2x = -2:

y=2(−2)3+3=2(−8)+3=−16+3=−13y = 2(-2)^3 + 3 = 2(-8) + 3 = -16 + 3 = -13

Step 4

State the equation of the tangent

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Answer

The point of tangency is (−2,−13)(-2, -13) with a slope of 24. We can use the point-slope form of the equation of a line:

y−y1=m(x−x1)y - y_1 = m(x - x_1) Substituting the values:

y−(−13)=24(x−(−2))y - (-13) = 24(x - (-2)) This simplifies to:

y+13=24(x+2)y + 13 = 24(x + 2) Thus,

y=24x+48−13⇒y=24x+35y = 24x + 48 - 13 \Rightarrow y = 24x + 35

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