Question Video: Finding the Value of an Unknown in a Quadratic Function given Its Graph | Nagwa Question Video: Finding the Value of an Unknown in a Quadratic Function given Its Graph | Nagwa

Question Video: Finding the Value of an Unknown in a Quadratic Function given Its Graph Mathematics

The function drawn below represents the quadratic 𝑓(π‘₯) = βˆ’π‘₯Β² + 3π‘₯ + π‘˜ βˆ’ 2, which intersects the π‘₯-axis at the points 𝐴 and 𝐡. Find the value of π‘˜ given 𝑂𝐡 = 4𝑂𝐴.

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Video Transcript

The function drawn below represents a quadratic 𝑓 of π‘₯ equals negative π‘₯ squared plus three π‘₯ plus π‘˜ minus two which intersects the π‘₯-axis at the points 𝐴 and 𝐡. Find the value of π‘˜ given 𝑂𝐡 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 𝑂𝐡 is equal to four times 𝑂𝐴. So let’s define the distance 𝑂𝐴 to be π‘Ž units. And therefore, the distance 𝑂𝐡 is four π‘Ž units. Now, what that actually means is that the coordinate 𝐴 can be represented as the point negative π‘Ž zero, whereas the coordinate 𝐡 must be four π‘Ž zero. So we’re essentially going to work backwards to find the value of π‘Ž and ultimately π‘˜. The roots of our equation are negative π‘Ž and four π‘Ž. So this means when we factor the expression for 𝑓 of π‘₯ and set it equal to zero, we should have something along the lines of negative π‘₯ minus π‘Ž times π‘₯ minus four π‘Ž.

Now, the reason I’ve chosen negative π‘₯ and π‘₯ is because I know the coefficient of π‘₯ squared to be negative one. And if I was going to go back and solve this equation, I would say that either negative π‘₯ minus π‘Ž equals zero or π‘₯ minus four π‘Ž equals zero. And when I solve each of these four π‘₯, I get π‘₯ equals negative π‘Ž, that’s our first route, and π‘₯ equals four π‘Ž; 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 π‘₯ minus π‘Ž times π‘₯ minus four π‘Ž. Let’s distribute these parentheses. We multiply the first term in each expression, negative π‘₯ times π‘₯ is negative π‘₯ squared. And we then multiply the outer terms, negative π‘₯ times negative four π‘Ž is four π‘₯π‘Ž.

Multiplying the inner terms gives us negative π‘₯π‘Ž and multiplying the last terms gives us four π‘Ž squared. We collect like terms, and we see that this simplifies to negative π‘₯ squared plus three π‘₯π‘Ž plus four π‘Ž squared. But of course, the question tells us that our function is equal to negative π‘₯ squared plus three π‘₯ plus π‘˜ minus two. So let’s equate coefficients and see if we can work out the value of π‘Ž and, therefore, the value of π‘˜. We’ll begin by equating the coefficients of π‘₯ or π‘₯ to the power of one. On the left-hand side, the coefficient of π‘₯ is three π‘Ž. And on the right-hand side, it’s three. If three π‘Ž is equal to three than π‘Ž 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 π‘₯ to the power of zero. On the left-hand side, we have four π‘Ž squared. And on the right-hand side, our constant term is π‘˜ minus two. Now, of course, we have just worked out that π‘Ž is equal to one. So we have four times one squared equals π‘˜ minus two or simply four equals π‘˜ minus two. We solve this equation, for π‘˜, by adding two to both sides. And we find π‘˜ to be equal to six. So given the function negative π‘₯ squared plus three π‘₯ plus π‘˜ minus two such that 𝑂𝐡 equals four 𝑂𝐴, we find π‘˜ to be equal to six.

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