Consider the graph. Which of the following could be the equation of the parabola? Is it A) 𝑦 equals 𝑥 plus one times 𝑥 plus five, B) 𝑦 equals negative 𝑥 plus one times 𝑥 plus five, C) 𝑦 equals negative 𝑥 plus one times 𝑥 minus five, D) 𝑦 equals 𝑥 minus one times 𝑥 minus five, or E) 𝑦 equals negative 𝑥 minus one times 𝑥 minus five?
This parabola is a quadratic graph. That means its equation has a power of 𝑥 no larger than two. We’re going to need decide which quadratic equation represents the equation of this graph. So, let’s consider what else we know about the shape of the curve. When a quadratic equation has a positive leading coefficient — that is, the coefficient of 𝑥 squared is positive — we have a U-shaped parabola. When, however, the coefficient of 𝑥 squared is negative — that is, 𝑦 equals negative 𝑥 squared, 𝑦 equals negative five 𝑥 squared plus two 𝑥, and so on — we have what we might call an inverted parabola as shown.
Now, we can see that the parabola on our curve is indeed inverted, so we know that the leading coefficient, in that case the coefficient of 𝑥 squared, is negative. And this actually means we can already disregard two of our equations. We’re going to disregard equation A, 𝑦 equals 𝑥 plus one times 𝑥 plus five, and equation D, 𝑦 equals 𝑥 minus one times 𝑥 minus five. And that’s because when we multiply two binomials of the form 𝑥 plus some constant 𝑎 and 𝑥 plus some constant 𝑏, we begin by multiplying the first term in each expression. So, that’s 𝑥 times 𝑥, which gives us the 𝑥 squared. Now, of course, we have three other terms when we multiply, but actually we get that positive leading coefficient of 𝑥 squared, so we know that we can disregard A and D.
B, C, and E are the product of two binomials as we saw before. But then, the entire expression is multiplied by negative one. So, we’re definitely going to have that negative leading coefficient. So, what next? Well, now we’ve considered the shape of the curve. We’re going to find the roots of each equation. That is, where does the graph cross the 𝑥-axis? And we achieve this by setting 𝑦 equal to zero. So, for graph B, we have negative 𝑥 plus one times 𝑥 plus five equals zero. We’re going to solve this for 𝑥. We can multiply through by negative one, and our equation becomes 𝑥 plus one times 𝑥 plus five equals zero.
Now, for the product of two expressions to be equal to zero, we absolutely know that at least one of those expressions must itself be zero. So, either 𝑥 plus one is equal to zero or 𝑥 plus five is equal to zero. When we subtract one from both sides of this first equation, we find 𝑥 is equal to negative one. And when we subtract five from both sides of our second equation, we get 𝑥 equals negative five. So, we know that the solutions to the equation negative 𝑥 plus one times 𝑥 plus five equals zero are 𝑥 equals negative one and 𝑥 equals negative five. Those are the points on the 𝑥-axis where the graph intersects.
Let’s repeat this and find the roots for graph C. That’s negative 𝑥 plus one times 𝑥 minus five equals zero. Once again, we can multiply by negative one. And this time we know that either 𝑥 plus one is equal to zero or 𝑥 minus five is equal to zero. When we solve these two equations, we get 𝑥 equals negative one and 𝑥 equals five. Let’s repeat this process one more time for graph E. We see that negative 𝑥 minus one times 𝑥 minus five equals zero. We multiply through by negative one, and this time we see that either 𝑥 minus one equals zero or 𝑥 minus five equals zero. Solving these equations, we find 𝑥 equals one and 𝑥 equals five to be the solutions to our quadratic equation.
Now remember, each pair of solutions tells us where the curve crosses the 𝑥-axis. If we go back to the parabola, we see that both points of intersection with the 𝑥-axis lie on the negative side. That means they absolutely must be negative one and negative five. This is the only equation for which both roots are indeed negative. And so, this means that graph B is the one we’re interested in. It’s 𝑦 equals negative 𝑥 plus one times 𝑥 plus five.