# Video: Finding the Equations of the Tangents to a Quadratic Curve at the Points of Intersection of the Curve with the 𝑥-Axis

Find the equations to the tangent lines of the curve 𝑦 = (𝑥 + 8)(𝑥 + 10) at the points where this curve intersects the 𝑥-axis.

05:05

### Video Transcript

Find the equations to the tangent lines of the curve 𝑦 equals 𝑥 plus eight multiplied by 𝑥 plus 10 at the points where this curve intersects the 𝑥-axis.

Let’s begin, first of all, by determining the coordinates of these points. We’re looking for the points where the curve intersects the 𝑥-axis. And we know that everywhere on the 𝑥-axis, 𝑦 is equal to zero. So by setting our equation for 𝑦 equal to zero, we have an equation which we can solve to find the 𝑥-coordinates of the points where the curve intersects the 𝑥-axis.

This is a quadratic equation already in its factored form. So we can take each factor in turn, set it equal to zero, and then solve the resulting linear equation, giving 𝑥 plus eight equals zero, which leads to 𝑥 equals negative eight and 𝑥 plus 10 equals zero, which leads to 𝑥 equals negative 10. The two points at which this curve intersects the 𝑥-axis then are the points negative 10, zero and negative eight, zero. We can sketch this curve, if we want. It’s a quadratic curve with a positive leading coefficient intersecting the 𝑥-axis at negative 10 and negative eight. So it looks a little something like this.

We’re then looking to find the equations of the lines, which are tangent to this curve at the points where it intersects the 𝑥-axis. So those are the two lines drawn in green. We know that the general equation of a straight line in its point-slope form is 𝑦 minus 𝑦 one equals 𝑚 𝑥 minus 𝑥 one, where 𝑥 one, 𝑦 one gives the coordinates of one point on the line and 𝑚 gives its slope. We know the coordinates of one point on each of our tangent lines. So in order to apply this formula, we need to determine their slopes. We can do this by recalling that the slope of the line tangent to a curve at any given point is the same as the slope of the curve itself at that point, which we can find by evaluating its first derivative.

Before we get into differentiating, let’s manipulate our expression for this curve slightly by distributing the parentheses and then simplifying to give 𝑦 is equal to 𝑥 squared plus 18𝑥 plus 80, a polynomial. We recall then that the power rule of differentiation, which tells us that the derivative with respect to 𝑥 of a general power term 𝑎𝑥 to the 𝑛th power for real values of 𝑎 and 𝑛 is equal to 𝑎𝑛 multiplied by 𝑥 to the power of 𝑛 minus one. We multiply by the exponent and then reduce the exponent by one.

We can now apply this power rule of differentiation to find the derivative of our function 𝑦. The derivative of 𝑥 squared with respect to 𝑥 is two 𝑥. The derivative of positive 18𝑥 is positive 18, which we can see if we think of 18𝑥 as 18𝑥 to the power of one. And finally, the derivative of a constant term positive 80 is just zero, which we can see if we think of 80 as 80𝑥 to the power of zero. We find then that the general slope function of our curve d𝑦 by d𝑥 is equal to two 𝑥 plus 18.

We then need to evaluate the slope at each of the points we’re interested in. First of all, when 𝑥 is equal to negative 10, the slope will be equal to two multiplied by negative 10 plus 18. That’s negative 20 plus 18 which is equal to negative two. Secondly, when 𝑥 is equal to negative eight, the slope will be equal to two multiplied by negative eight plus 18. That’s negative 16 plus 18 which is equal to two. Notice that this is consistent with what we’ve seen on our sketch. The tangent line at the point where 𝑥 equals negative 10 has a negative slope. Whereas the tangent to the point when 𝑥 equals negative eight has positive slope.

Next, we can use the general equation of our straight line. At the point with coordinates negative 10, zero where the slope is equal to negative two, we have the equation 𝑦 minus zero equals negative two multiplied by 𝑥 minus negative 10. That simplifies to 𝑦 equals negative two multiplied by 𝑥 plus 10. And distributing the parentheses, we have 𝑦 equals negative two 𝑥 minus 20. We can then collect all the terms on the left-hand side of the equation to give the equation of this tangent as 𝑦 plus two 𝑥 plus 20 is equal to zero.

For our other tangent at the point negative eight, zero where the slope is equal to two, we have the equation 𝑦 minus zero equals two multiplied by 𝑥 minus negative eight. That simplifies to 𝑦 equals two multiplied by 𝑥 plus eight. And then distributing the parentheses gives 𝑦 equals two 𝑥 plus 16. Again, collecting all terms on the left-hand side of the equation gives the equation of this tangent line in the form 𝑦 minus two 𝑥 minus 16 is equal to zero.

So we’ve completed the problem and found the equations of both lines, which are tangent to this curve at the points where it intersects the 𝑥-axis. They are 𝑦 plus two 𝑥 plus 20 equals zero and 𝑦 minus two 𝑥 minus 16 equals zero.