# Question Video: Identifying Graphs of Quadratic Equations in Factored Form Mathematics

Which of the following graphs represents the equation 𝑓(𝑥) = (𝑥 + 4)(𝑥 − 2)? [A] Graph A [B] Graph B [C] Graph C [D] Graph D [E] Graph E

03:25

### Video Transcript

Which of the following graphs represents the equation 𝑓 of 𝑥 equals 𝑥 plus four times 𝑥 minus two? Options A, B, C, D, and E.

In this question, we are given a function 𝑓 of 𝑥 and asked to determine which of the five given graphs represent this function. We can note that 𝑓 of 𝑥 is the product of two linear factors, and if we were to expand this expression, we would obtain a quadratic.

We could answer this question by eliminating options; however, it is a more useful skill to be able to sketch quadratics from their equations. To do this, we first note that for a product to be equal to zero, one of the factors must be equal to zero. This means that the function 𝑓 of 𝑥 can only be equal to zero when 𝑥 equals negative four or 𝑥 equals two. If the function outputs zero at these values of 𝑥, then they must be the 𝑥-intercepts of its graph. This is a good starting point to begin our sketch, but we can determine more information about the graph of this function by expanding the product.

We can do this by finding the sum of the product of each pair of terms from the two factors. We find that 𝑓 of 𝑥 is equal to 𝑥 squared plus four 𝑥 minus two 𝑥 minus eight. This gives us 𝑥 squared plus two 𝑥 minus eight. We now see that 𝑓 of 𝑥 is a quadratic equation with leading coefficient one and constant term negative eight.

We can then recall that if a quadratic function has a positive leading coefficient, then its graph will be a parabola that opens upwards. In other words, it is U-shaped. We can also recall that the constant term in the function tells us the value of the 𝑦-intercept, since we can substitute 𝑥 equals zero into the function to be left with only the constant term.

This is now enough information to sketch the parabola. First, we can add the 𝑥-intercepts at negative four and two. Second, we can add the 𝑦-intercept at negative eight onto our sketch. Finally, we connect these points with a parabolic shape that opens upwards to get the following sketch. We can now compare this sketch to each of the five given options.

In option (A), we can note there are no 𝑥-intercepts and the 𝑦-intercept is not at negative eight, so this is not the correct graph. In option (B), we can see from the graph that the 𝑦-intercept is not at negative eight and that the parabola opens downwards, not upwards, so this cannot be the correct graph. In option (C), we see that the 𝑥-intercepts are not at negative four and two, and the 𝑦-intercept is not at negative eight. So, this is not the correct graph.

In option (D), we can see that the 𝑥-intercepts are not at negative four and two. So, this cannot be the correct graph. Finally, in option (E), we can see that 𝑥- and 𝑦-intercepts are in the correct places and we also have a parabolic curve which opens upwards, so this must be the correct option.

Hence, the graph of 𝑓 of 𝑥 equals 𝑥 plus four times 𝑥 minus two is given by option (E).