Video: Finding Unknown Coefficients in a Polynomial Function given the Equation of Its Tangent at a Given Point

If the curve 𝑦 = π‘Žπ‘₯Β³ + 𝑏π‘₯Β² + 2π‘₯ + 7 is tangent to the line 𝑦 = 7π‘₯ βˆ’ 3 at (βˆ’1, βˆ’10), find the constants π‘Ž and 𝑏.

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Video Transcript

If the curve 𝑦 is equal to π‘Žπ‘₯ cubed plus 𝑏π‘₯ squared plus two π‘₯ plus seven is tangent to the line 𝑦 is equal to seven π‘₯ minus three at the point negative one, negative 10, find the constants π‘Ž and 𝑏.

The question tells us that our cubic curve is tangent to our line 𝑦 is equal to seven π‘₯ minus three at the point negative one, negative 10. We need to use this information to find the values of our constants π‘Ž and 𝑏. To start, let’s recall what it means for our curve to be tangent to our line at the point negative one, negative 10.

For our curve to be tangent to our line at the point negative one, negative 10, our curve must pass through the point negative one, negative 10. Otherwise, our curve and our line will not touch at this point. Next, we need the slope of our curve and the slope of our line to be equal at the point negative one, negative 10. So, we have two conditions which must be satisfied for our curve to be tangent to the line at the point negative one, negative 10.

Let’s start by finding all of the values of π‘Ž and 𝑏 such that our curve passes through the point negative one, negative 10. To check if our curve is passing through the point negative one, negative 10, we’ll just substitute π‘₯ is equal to negative one and 𝑦 is equal to negative 10 into the equation of our curve. Substituting π‘₯ is equal to negative one and 𝑦 is equal to negative 10 into the equation of our curve gives us negative 10 is equal to π‘Ž times negative one cubed plus 𝑏 times negative one squared plus two times negative one plus seven.

We can simplify this to give us negative 10 is equal to negative π‘Ž plus 𝑏 plus five. And we’ll subtract five from both sides of this equation to get negative 15 is equal to negative π‘Ž plus 𝑏. So, all values of π‘Ž and 𝑏 which satisfy this equation will mean that our curve passes through the point negative one, negative 10. Let’s now find all of the values of π‘Ž and 𝑏 such that the slope of our curve will be equal to the slope of our line at the point negative one, negative 10.

Let’s start by finding the slope of our line. We recall the coefficient of π‘₯ in our equation for the line is just the slope of our line, which in this case is seven. So, we know the slope of our line is equal to seven. To find the slope of our curve, we’re going to need to differentiate it with respect to π‘₯. This gives us the derivative of π‘Žπ‘₯ cubed plus 𝑏π‘₯ squared plus two π‘₯ plus seven with respect to π‘₯.

And since this is a polynomial, we can differentiate each term by using the power rule for differentiation. We multiply by the exponent of π‘₯ and reduce this exponent by one. Applying this to each of our terms, we get d𝑦 by dπ‘₯ is equal to three π‘Žπ‘₯ squared plus two 𝑏π‘₯ plus two. Remember, we want the slope of our curve at the point negative one, negative 10 to be equal to seven. So, when π‘₯ is equal to negative one, our slope is equal to seven.

We’ll substitute π‘₯ is equal to negative one into our equation for the slope. Substituting π‘₯ is equal to negative one and the slope of seven gives us seven is equal to three π‘Ž times negative one squared plus two 𝑏 times negative one plus two. And if we simplify and rearrange this equation, we get that five is equal to three π‘Ž minus two 𝑏. And any pairs of values for π‘Ž and 𝑏 which satisfy this equation will mean that the slope of our curve when π‘₯ is equal to negative one is seven.

So, for both of our conditions for our curve to be a tangent of the line at this point to be true, we need both of these equations to be true. In other words, to find the values of π‘Ž and 𝑏, we need to solve these two equations as simultaneous equations. There’s a few different ways of doing this.

We’re going to rearrange the equation negative 15 is equal to negative π‘Ž plus 𝑏 to make 𝑏 the subject. To do this, we just add π‘Ž to both sides of the equation. We get π‘Ž minus 15 is equal to 𝑏. All we have to do now is substitute this expression for 𝑏 into our other simultaneous equation. Substituting 𝑏 is equal to π‘Ž minus 15, we get five is equal to three π‘Ž minus two times π‘Ž minus 15. Distributing negative two over our parentheses gives us five is equal to three π‘Ž minus two π‘Ž plus 30.

But three π‘Ž minus two π‘Ž is equal to π‘Ž. So, we just subtract 30 from both sides of this equation. And we see that π‘Ž is equal to negative 25. Now, we just need to find the value of 𝑏. We’ll substitute π‘Ž is equal to negative 25 into our equation for 𝑏. Substituting π‘Ž is equal to negative 25, we get 𝑏 is equal to negative 25 minus 15, which tells us that 𝑏 is equal to negative 40. Therefore, we’ve shown for the curve 𝑦 is equal to π‘Žπ‘₯ cubed plus 𝑏π‘₯ squared plus two π‘₯ plus seven. To be tangent to the line 𝑦 is equal to seven π‘₯ minus three at the point negative one, negative 10. Then π‘Ž must be equal to negative 25 and 𝑏 must be equal to negative 40.

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