Video: Matching a Quadratic Equation to Its Graph

Which of the following could be the equation graphed in the given figure? [A] 𝑦 = (4π‘₯ βˆ’ 1)(4π‘₯ + 3) [B] 𝑦 = 3(4π‘₯ + 1)(3π‘₯ βˆ’ 4) [C] 𝑦 = 2(π‘₯ + 4)(4π‘₯ βˆ’ 3) [D] 𝑦 = 3(π‘₯ βˆ’ 4)(4π‘₯ + 3)


Video Transcript

Which of the following could be the equation graphed in the given figure? A) 𝑦 equals four π‘₯ minus one times four π‘₯ plus three. B) 𝑦 equals three times four π‘₯ plus one times three π‘₯ minus four. C) 𝑦 equals two times π‘₯ plus four times four π‘₯ minus three. Or D) 𝑦 equals three times π‘₯ minus four times four π‘₯ plus three.

In this image, we see a curve that opens upward. We know that this is a quadratic equation. If we’re given a quadratic in the form 𝑦 equals π‘Žπ‘₯ squared plus 𝑏π‘₯ plus 𝑐, when π‘Ž is positive, the graph opens upward. And if π‘Ž has a negative value, the graph opens facing downward. In addition to that, we can say that 𝑐 is the 𝑦-intercept. It’s the place that graph crosses when π‘₯ equals zero. At this point, none of these four equations are given to us in the form π‘Žπ‘₯ squared plus 𝑏π‘₯ plus 𝑐. These four equations have already been factored. But it’s helpful to identify the π‘Ž-variable to check and see if we can eliminate any of the four choices.

The π‘Ž-variable is the coefficient of the π‘₯ squared term, which means it’s found when we multiply the two π‘₯ terms together. Here, we have four π‘₯ times four π‘₯, which gives us 16π‘₯ squared. For option B, the coefficient of π‘₯ squared would be three times four π‘₯ times three π‘₯, 36π‘₯ squared. And for C, two times π‘₯ times four π‘₯, which is eight π‘₯ squared. And for option D, three times π‘₯ times four π‘₯ equals 12π‘₯ squared. In this case, all four of the equations have a positive π‘Ž-value. So we can’t eliminate any of the four.

In our next step, let’s consider the 𝑐-value, the 𝑦-intercept. On this graph, the 𝑦-intercept is at negative three. It is the point zero, negative three. In the correct equation, if you plug in zero for π‘₯, you should get a 𝑦-value of negative three. In our first equation, equation A, if we plugged in zero for π‘₯, we would have four times zero, which is zero. Zero minus one is negative one, and zero plus three is positive three. And so, for equation one, 𝑦 is equal to negative one times three. Equation one has a 𝑐-value of negative three. And we can say that this could be our equation.

Following the same procedure, if we plug in zero for π‘₯ in option B, we’ll have zero plus one times zero minus four. And we can’t forget this constant of three. 𝑦 will be equal to three times one times negative four when π‘₯ is equal to zero. 𝑐 equals negative 12 here, which means 𝑏 is not the graphed equation. Following the same procedure, plug in zero for π‘₯. Zero plus four is four. Zero minus three is negative three. So we have a 𝑐-value of two times four times negative three, which is negative 24. And that means that 𝑐 can’t be our equation. Finally, we do this one more time. Plug in zero. Zero minus four is negative four. Zero plus three is three. When π‘₯ is zero, 𝑦 is equal to three times negative four times three, which is negative 36.

Of these four equations, only one of them had a correct 𝑦-intercept of negative three. And so, we think that the equation 𝑦 equals four π‘₯ minus one times four π‘₯ plus three is our equation. There are two more points on this line we can use to check and make sure this is true. On the graph, we see that there are four marks between zero and one. And that means we have an intersection at the value π‘₯ equals one-fourth. When π‘₯ equals one-fourth, 𝑦 equals zero. It crosses the-π‘₯ axis. The same thing is true in the other direction. We have four marks between zero and negative one. And our intersection is at the third mark to the left, which is π‘₯ equals negative three-fourths. When π‘₯ equals negative three-fourths, 𝑦, again, equals zero.

When we have a factored equation like this, we can set each factor equal to zero for π‘₯ minus one equals zero and for π‘₯ plus three equals zero. Doing this helps us find where the equation crosses the π‘₯-axis. On the left, we add one to both sides. We then have an equation that says four π‘₯ equals one. And then we divide by four on both sides, which tells us when π‘₯ equals one-fourth, 𝑦 equals zero. And it verifies one of the solutions on our graph.

So on the other side, we subtract three from both sides. And we get four π‘₯ equals negative three. From there, we divide both sides by four. Because four π‘₯ divided by four equals one, we can say that π‘₯ is equal to negative three over four. And when π‘₯ is negative three-fourths, 𝑦 equals zero. Which confirms the second π‘₯-intercept. And so, we say that the equation graphed in this figure could be 𝑦 equals four π‘₯ minus one times four π‘₯ plus three.

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