Video Transcript
The function drawn below represents a quadratic 𝑓 of 𝑥 equals negative 𝑥 squared plus three 𝑥 plus 𝑘 minus two which intersects the 𝑥-axis at the points 𝐴 and 𝐵. Find the value of 𝑘 given 𝑂𝐵 is four times 𝑂𝐴.
We have a quadratic function with a negative leading coefficient. That gives us the inverted parabola shape shown. We’re also told that 𝑂𝐵 is equal to four times 𝑂𝐴. So let’s define the distance 𝑂𝐴 to be 𝑎 units. And therefore, the distance 𝑂𝐵 is four 𝑎 units. Now, what that actually means is that the coordinate 𝐴 can be represented as the point negative 𝑎 zero, whereas the coordinate 𝐵 must be four 𝑎 zero. So we’re essentially going to work backwards to find the value of 𝑎 and ultimately 𝑘. The roots of our equation are negative 𝑎 and four 𝑎. So this means when we factor the expression for 𝑓 of 𝑥 and set it equal to zero, we should have something along the lines of negative 𝑥 minus 𝑎 times 𝑥 minus four 𝑎.
Now, the reason I’ve chosen negative 𝑥 and 𝑥 is because I know the coefficient of 𝑥 squared to be negative one. And if I was going to go back and solve this equation, I would say that either negative 𝑥 minus 𝑎 equals zero or 𝑥 minus four 𝑎 equals zero. And when I solve each of these four 𝑥, I get 𝑥 equals negative 𝑎, that’s our first route, and 𝑥 equals four 𝑎; that’s our second route as required. So what we can say then is that our function, when factored, must look like this. It’s negative 𝑥 minus 𝑎 times 𝑥 minus four 𝑎. Let’s distribute these parentheses. We multiply the first term in each expression, negative 𝑥 times 𝑥 is negative 𝑥 squared. And we then multiply the outer terms, negative 𝑥 times negative four 𝑎 is four 𝑥𝑎.
Multiplying the inner terms gives us negative 𝑥𝑎 and multiplying the last terms gives us four 𝑎 squared. We collect like terms, and we see that this simplifies to negative 𝑥 squared plus three 𝑥𝑎 plus four 𝑎 squared. But of course, the question tells us that our function is equal to negative 𝑥 squared plus three 𝑥 plus 𝑘 minus two. So let’s equate coefficients and see if we can work out the value of 𝑎 and, therefore, the value of 𝑘. We’ll begin by equating the coefficients of 𝑥 or 𝑥 to the power of one. On the left-hand side, the coefficient of 𝑥 is three 𝑎. And on the right-hand side, it’s three. If three 𝑎 is equal to three than 𝑎 must be equal to one. And of course, we could solve this equation more formally by dividing both sides by three.
We’ll now equate the constants. That’s the coefficient of 𝑥 to the power of zero. On the left-hand side, we have four 𝑎 squared. And on the right-hand side, our constant term is 𝑘 minus two. Now, of course, we have just worked out that 𝑎 is equal to one. So we have four times one squared equals 𝑘 minus two or simply four equals 𝑘 minus two. We solve this equation, for 𝑘, by adding two to both sides. And we find 𝑘 to be equal to six. So given the function negative 𝑥 squared plus three 𝑥 plus 𝑘 minus two such that 𝑂𝐵 equals four 𝑂𝐴, we find 𝑘 to be equal to six.
