Question Video: Finding the Unknown Coefficients in a Quadratic Function given the Slope of the Tangent to Its Curve at a Point Mathematics

Video Transcript

Let 𝑓 of 𝑥 equal 𝑎𝑥 squared plus 𝑏 with constants 𝑎, 𝑏. Determine the constants for which the slope of the tangent to the curve 𝑦 equals 𝑓 of 𝑥 at negative two, 29 is negative 24.

We need to determine the values of two unknown constants 𝑎 and 𝑏, which are used in the definition of this function 𝑓 of 𝑥. To do this, we’ll need to use the information given in the question to form and then solve some equations. First, if there is a tangent to the curve 𝑦 equals 𝑓 of 𝑥 at the point with coordinates negative two, 29, then the point with these coordinates lies on the curve. In other words, when 𝑥 is equal to negative two, 𝑓 of 𝑥 is equal to 29. And so we can use this to find our first equation connecting 𝑎 and 𝑏.

Substituting 𝑥 equals negative two and 𝑓 of 𝑥 equals 29 into the definition of 𝑓 of 𝑥 gives 29 equals 𝑎 multiplied by negative two squared plus 𝑏. This simplifies to 29 equals four 𝑎 plus 𝑏. We can’t solve this equation as it is a single equation with two unknowns. So let’s consider the other information we were given in the question, which is that the slope of this tangent is negative 24.

Now, a tangent to a curve at any given point has the same slope as the curve itself at that point. The slope of a curve at any given point, which is also its instantaneous rate of change at that point, can be found by evaluating its first derivative at that point. The first derivative of 𝑓 of 𝑥 is denoted as 𝑓 prime of 𝑥. So we need to find an expression for 𝑓 prime of negative two, because the 𝑥-value at this point is negative two.

To find the derivative of 𝑓 of 𝑥, which is a polynomial, we can recall the power rule of differentiation. This tells us that the derivative with respect to 𝑥 of 𝑎 multiplied by 𝑥 to the power of 𝑛, where 𝑎 and 𝑛 are real constants, is 𝑎𝑛 multiplied by 𝑥 to the power of 𝑛 minus one. We multiply by the exponent, and then we reduce the exponent by one. To find the derivative of the first term in 𝑓 of 𝑥, which is 𝑎𝑥 squared, we multiply by the exponent of two and we reduce the exponent by one, giving two 𝑎𝑥 to the first power. But of course 𝑥 to the first power is simply 𝑥.

The derivative of a constant with respect to 𝑥 is simply zero, which we can see if we think of the constant 𝑏 as 𝑏 multiplied by 𝑥 to the power of zero. When we apply the power rule of differentiation, we get zero multiplied by 𝑏 multiplied by 𝑥 to the power of negative one. But of course zero multiplied by anything is simply zero. So we find that 𝑓 prime of 𝑥 is equal to two 𝑎𝑥.

We know that when 𝑥 is equal to negative two, the slope of the tangent — and hence the first derivative of the function — is equal to negative 24. So we have the equation negative 24 is equal to two multiplied by negative two multiplied by 𝑎. That’s negative 24 is equal to negative four 𝑎. And to solve for 𝑎, we can divide each side of this equation by negative four, giving 𝑎 is equal to six.

So we found the value of 𝑎. And all that remains is to find the value of 𝑏. We can do this by substituting 𝑎 equals six back into the first equation we found relating 𝑎 and 𝑏. This gives four multiplied by six plus 𝑏 is equal to 29. Four multiplied by six is 24. So we have 24 plus 𝑏 equals 29. And then subtracting 24 from each side, we find that 𝑏 is equal to five. The equation of the curve then 𝑓 of 𝑥 is 𝑓 of 𝑥 equals six 𝑥 squared plus five. But we’re asked just to give the values of the constants 𝑎 and 𝑏. Our answer is that 𝑎 is equal to six and 𝑏 is equal to five.