Video: Finding the Equation of the Tangent to the Curve of a Polynomial Function at a Given Point

Determine the equation of the line tangent to the curve 𝑦 = 4π‘₯Β³ βˆ’ 2π‘₯Β² + 4 at the point (βˆ’1, βˆ’2).

02:16

Video Transcript

Determine the equation of the line tangent to the curve 𝑦 equals four π‘₯ cubed minus two π‘₯ squared plus four at the point negative one, negative two.

So we’ve been given the equation of a curve. And we need to determine the equation of the line that is tangent to this curve at a particular point. We’re going to use the formula for the general equation of a straight line 𝑦 minus 𝑦 one equals π‘š π‘₯ minus π‘₯ one. We already know the coordinates π‘₯ one, 𝑦 one. It’s the point negative one, negative two. But what about π‘š, the slope of this line? Well, we recall that the slope of a curve is equal to the slope of the tangent to the curve at that point. So in order to find the slope of this tangent, we’re first going to find the gradient function of the curve d𝑦 by dπ‘₯.

We can do this by applying the power rule of differentiation, giving d𝑦 by dπ‘₯ equals four multiplied by three π‘₯ squared minus two multiplied by two π‘₯. Remember, a constant differentiates to zero. So that plus four just differentiates to zero in our derivative, which simplifies to 12π‘₯ squared minus four π‘₯. Now, that’s the general gradient function of this curve. But we want to know the gradient at a particular point. So we need to evaluate d𝑦 by dπ‘₯ when π‘₯ is equal to negative one because that’s our π‘₯-coordinate at this point. This gives 12 multiplied by negative one squared minus four multiplied by negative one, which simplifies to 16.

We now know that the slope of this tangent is 16 and the coordinates of a point that it passes through are negative one, negative two. So we have all the information we need in order to use the formula for the general equation of a straight line. Substituting the values of π‘š, π‘₯ one, and 𝑦 one gives 𝑦 minus negative two equals 16 π‘₯ minus negative one. That’s 𝑦 plus two equals 16π‘₯ plus 16. And, then, subtracting two from each side in order to collect the constants gives 𝑦 equals 16π‘₯ plus 14. So this is the equation of the line tangent to the given curve at the point negative one, negative two. It passes through the point negative one, negative two and it has the same gradient as the curve at that point.

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