# Video: Finding the Unknown Coefficients in the Expression of a Quadratic Function given a Point on Its Curve and the Value of the Derivative at This Point

Find all pairs of values for 𝑎 and 𝑏 so that the point (3,1) lies on the curve 𝑦 = (𝑎𝑥 + 𝑏)² and d𝑦/d𝑥 = 6 at this point.

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### Video Transcript

Find all pairs of values for 𝑎 and 𝑏 so that the point three, one lies on the curve 𝑦 is equal to 𝑎𝑥 plus 𝑏 all squared and d𝑦 by d𝑥 is equal to six at this point.

The question wants us to find every possible pair of values for 𝑎 and 𝑏 so that the point three, one lies on our curve 𝑦 is equal to 𝑎𝑥 plus 𝑏 squared and the slope of the curve is equal to six at this point. So we’re actually given two pieces of information about our curve. We’re told that point three, one lies on our curve, and we’re told the slope of our curve is six at this point. Let’s use each piece of information separately.

First, let’s find all possible values of 𝑎 and 𝑏 such that the point three, one lies on our curve. We know saying three, one lies on our curve is the same as saying that our equation holds when we substitute in 𝑥 is equal to three and 𝑦 is equal to one. We’ll substitute these values in. This gives us one is equal to 𝑎 times three plus 𝑏 all squared. And we’ll rewrite 𝑎 times three as three 𝑎. Finally, we’ll distribute the square over our parentheses. This gives us the equation one is equal to nine 𝑎 squared plus six 𝑎𝑏 plus 𝑏 squared. Let’s recap what this equation means.

Any pair of values for 𝑎 and 𝑏, which satisfy this equation, will mean that the point three, one lies on our curve. We can form a similar equation to make the slope of our curve equal to six at the point three, one. The first thing we need to do is find an equation for our slope. And we’ll do this by distributing the exponent over our parentheses. This gives us 𝑦 is equal to 𝑎 squared 𝑥 squared plus two 𝑎𝑏𝑥 plus 𝑏 squared. We want to use this to find an expression for d𝑦 by d𝑥. We can do this by differentiating each term by using the power rule for differentiation. Which tells us, for real constants 𝑐 and 𝑛, the derivative of 𝑐𝑥 to the 𝑛th power with respect to 𝑥 is equal to 𝑛 times 𝑐 times 𝑥 to the power of 𝑛 minus one. We multiply by the exponent of 𝑥 and reduce this exponent by one.

So let’s start by differentiating 𝑎 squared 𝑥 squared. We’ll multiply by the exponent of 𝑥, which is two, and then reduce this exponent by one. And we can simplify this. Two minus one is equal to one. And 𝑥 to the first power is just equal to 𝑥. To differentiate our second term, we’ll rewrite two 𝑎𝑏𝑥 as two 𝑎𝑏𝑥 times 𝑥 to the first power. Now, we can differentiate this by using the power rule for differentiation. We’ll multiply by the exponent of 𝑥, which is one, and reduce this exponent by one. And we can then simplify this since one minus one is equal to zero. And 𝑥 to the zeroth power is just equal to one. So this term is just two 𝑎𝑏.

Finally, we could differentiate the constant 𝑏 squared by writing it as 𝑏 squared times 𝑥 to the zeroth power. However, we can also just remember the derivative of any constant is just equal to zero. So this gives us an expression for d𝑦 by d𝑥. And remember, we know the slope is equal to six when 𝑥 is equal to three. So we’ll substitute this information in. This gives us six is equal to two 𝑎 squared times three plus two 𝑎𝑏. And we can simplify this equation by dividing through by two. And this gives us the equation three is equal to three 𝑎 squared plus 𝑎𝑏. And remember, any pair of values of 𝑎 and 𝑏 which satisfy this equation will mean the slope of our curve is equal to six at the point three, one.

Now, since we’re looking for the pairs of values of 𝑎 and 𝑏, which satisfy both of these properties, they must satisfy both of these equations. This means we need to solve these as simultaneous equations. There’s a few different ways of doing this. We’re going to rewrite the equation three is equal to three 𝑎 squared plus 𝑎𝑏 to make 𝑏 the subject. We’ll start by subtracting three 𝑎 squared from both sides. This gives us three minus three 𝑎 squared is equal to 𝑎𝑏. Next, we’ll divide both sides of our equation through by 𝑎. And this gives us 𝑏 is equal to three minus three 𝑎 squared all divided by 𝑎.

We now want to substitute this value of 𝑏 into our other simultaneous equation. This will give us an equation for all of the values of 𝑎 which satisfy both of our simultaneous equations. Substituting 𝑏 is equal to three minus three 𝑎 squared over 𝑎 gives us one is equal to nine 𝑎 squared plus six 𝑎 times three minus three 𝑎 squared over 𝑎 plus three minus three 𝑎 squared over 𝑎 squared. At this point, we want to cancel 𝑎 divided by 𝑎. However, we can only do this if 𝑎 is not equal to zero. So we need to check if 𝑎 is equal to zero. Let’s take a look at the equation of our curve when 𝑎 is equal to zero.

When 𝑎 is equal to zero, the equation of our curve becomes 𝑦 is equal to 𝑏 squared. This is just a horizontal line. So in particular, its slope is equal to zero for all values of 𝑥. But we want the slope to be equal to six at the point three, one. So 𝑎 cannot be equal to zero. So we can cancel 𝑎 divided by 𝑎. Next, we’ll distribute six over our parentheses. This gives us 18 minus 18𝑎 squared. Now, we’ll distribute the square over the parentheses in our last term. This gives us nine minus 18𝑎 squared plus nine 𝑎 to the fourth power divided by 𝑎 squared.

We’ll simplify this slightly. We’ll subtract one from both sides of our equation. And 18 minus one is 17. And we have nine 𝑎 squared minus 18𝑎 squared is equal to negative nine 𝑎 squared. So our equation is now zero is equal to 17 minus nine 𝑎 squared plus nine minus 18𝑎 squared plus nine 𝑎 to the fourth power divided by 𝑎 squared. We’ll simplify this by multiplying both sides of our equation through by 𝑎 squared. And remember, we know that 𝑎 is not equal to zero.

Multiplying each term by 𝑎 squared and simplifying, we get zero is equal to 17𝑎 squared minus nine 𝑎 to the fourth power plus nine minus 18𝑎 squared plus nine 𝑎 to the fourth power. And we can now simplify this. Negative nine 𝑎 to the fourth power plus nine 𝑎 to the fourth power is equal to zero. And 17𝑎 squared minus 18𝑎 squared is equal to negative 𝑎 squared. So we have zero is equal to nine minus 𝑎 squared. We’ll add 𝑎 squared to both sides of this equation to see that 𝑎 squared is equal to nine. And we can solve this for 𝑎 by taking the square root, remembering we get a positive and a negative result. This means that 𝑎 is equal to positive or negative three.

And we can now find the corresponding value of 𝑏 by substituting these values of 𝑎 into our equation for 𝑏. When 𝑎 is equal to three, we get that 𝑏 is equal to three minus three times three squared all divided by three, which we can calculate to give us negative eight. And when 𝑎 is equal to negative three, we get 𝑏 is equal to three minus three times negative three squared divided by negative three, which we can calculate to give us eight.

So we’ve shown if 𝑎 and 𝑏 satisfy both of these conditions, then 𝑎 is equal to three and 𝑏 is equal to negative eight, or 𝑎 is equal to negative three and 𝑏 is equal to positive eight. We’ll write this as 𝑎 is equal to positive or negative three and then 𝑏 is equal to negative or positive eight.