# Question Video: Finding the Equation of the Tangent to the Curve of a Quadratic Function at a Given Point

Find the equation of the tangent to the curve 𝑦 = 8𝑥² + 5𝑥 − 6 at the point (−1, −3).

03:19

### Video Transcript

Find the equation of the tangent to the curve 𝑦 equals eight 𝑥 squared plus five 𝑥 minus six at the point negative one, negative three.

So, we’ve been given the equation of a curve, it is a quadratic curve, and asked to find the equation of the tangent to the curve at a particular point. We recall, first of all, that the tangent to a curve at a given point passes through that point and has the same slope as the curve at that point. A tangent is a straight line. So, we can use the general form of the equation of a straight line, 𝑦 minus 𝑦 one equals 𝑚𝑥 minus 𝑥 one, in order to find its equation.

Here, 𝑚 represents the slope of the tangent, and the point 𝑥 one, 𝑦 one is a point on the line. We know that the point with coordinates negative one, negative three lies on this line because we’re looking for the tangent to the given curve at this point. And so, substituting negative one for 𝑥 one and negative three for 𝑦 one gives 𝑦 minus negative three equals 𝑚𝑥 minus negative one. This can of course be simplified to 𝑦 plus three equals 𝑚𝑥 plus one.

Now, we need to determine the value of 𝑚, the slope of the tangent. And recall that this is the same as the slope of the curve at this point. We can therefore use differentiation in order to find the slope of both the tangent and the curve. 𝑦 is a polynomial function of 𝑥, so we can use the power rule for differentiation in order to find its derivative d𝑦 by d𝑥. We recall that to find the derivative of a power of 𝑥, we multiply by that power and then decrease the power by one. So, we have that d𝑦 by d𝑥 is equal to eight multiplied by two 𝑥 plus five multiplied by one. And then, the derivative of a constant, negative six in this case, is just zero. Our expression for d𝑦 by d𝑥 therefore simplifies to 16𝑥 plus five.

Now, this is a general function for the slope at any point on the curve. We need to evaluate it at the point we’re interested in. That’s the point where 𝑥 is equal to negative one. Substituting negative one for 𝑥 gives that the derivative of 𝑦 with respect to 𝑥 at this point is equal to 16 multiplied by negative one plus five, which is equal to negative 11.

So, we now know the slope of the curve, and hence the slope of the tangent to the curve at this point. The final step is to substitute this value of 𝑚 into the equation of our tangent. Doing so gives 𝑦 plus three equals negative 11 multiplied by 𝑥 plus one. We can distribute the parentheses on the right-hand side to give 𝑦 plus three equals negative 11𝑥 minus 11. And then collect all the terms on the left-hand side of the equation to give 𝑦 plus 11𝑥 plus 14 is equal to zero.

So, by using differentiation to find the slope of the curve and hence the slope of the tangent to the curve at this point. And also recalling the general form of the equation of a straight line. We’ve found that the equation of the tangent to the curve 𝑦 equals eight 𝑥 squared plus five 𝑥 minus six at the point negative one, negative three is 𝑦 plus 11𝑥 plus 14 is equal to zero.