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Question Video: Identifying Graphs of Factorable Quadratic Equations Mathematics • Third Year of Preparatory School

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Which of the following graphs represents the equation ๐‘“(๐‘ฅ) = โˆ’2๐‘ฅยฒ + 9๐‘ฅ โˆ’ 7? [A] Graph A [B] Graph B [C] Graph C [D] Graph D [E] Graph E

02:59

Video Transcript

Which of the following graphs represents the equation ๐‘“ of ๐‘ฅ equals negative two ๐‘ฅ squared plus nine ๐‘ฅ minus seven?

There are two ways we can answer this question. But the first thing that weโ€™re going to do is establish whether the graph of our function opens upwards or downward. Firstly, we know that if we have the equation ๐‘“ of ๐‘ฅ equals ๐‘Ž๐‘ฅ squared plus ๐‘๐‘ฅ plus ๐‘ and the value of ๐‘Ž is positive, the graph opens upward. If itโ€™s negative, it opens downward.

Now, for our function, the value of ๐‘Ž is negative two. So itโ€™s negative. This means the function opens downward. And we can therefore disregard any functions that open upward. Thatโ€™s the function represented by graph (B) and the one represented by graph (E).

Now that weโ€™ve done this, weโ€™re going to plot a table of values to identify the points through which this graph passes. Weโ€™ll need to do this for a variety of values of ๐‘ฅ. Letโ€™s start with ๐‘ฅ equals zero. Thatโ€™s ๐‘“ of zero equals negative two times zero squared plus nine times zero minus seven. And thatโ€™s negative seven. So when ๐‘ฅ is equal to zero, the value of the function is negative seven. And that actually means that the ๐‘ฆ-intercept of our function is negative seven.

But of course the other technique we couldโ€™ve used is to identify that for the function ๐‘Ž๐‘ฅ squared plus ๐‘๐‘ฅ plus ๐‘, the ๐‘ฆ-intercept has coordinates zero, ๐‘. Either technique here, substituting or identifying the feature of the graph, is equally acceptable.

Weโ€™re now going to choose some positive and negative values of ๐‘ฅ to satisfy each of the graphs weโ€™ve been given. First, letโ€™s substitute ๐‘ฅ equals negative three into the function. We get negative two times negative three squared plus nine times negative three minus seven. Thatโ€™s negative 52. Similarly, letโ€™s substitute ๐‘ฅ equals negative two. When we do, we find that ๐‘“ of negative two is negative 33. And that means our graph must pass through the point negative two, negative 33. Substituting the remaining values, and we see that ๐‘“ of negative one is negative 18, ๐‘“ of one is zero, ๐‘“ of two is three, and ๐‘“ of three is two. Plotting the values that we can on our graphs, and we see we disregard option (C) completely. We notice itโ€™s the wrong side of the ๐‘ฆ-axis. And in fact the graph that does pass through these points is graph (A).

Now, in fact, there was an alternative method here. We began by identifying the shape of the graph. We saw that it opened downward. And we found the value of its ๐‘ฆ-intercept. We could also have set ๐‘“ of ๐‘ฅ equal to zero to find the value of the ๐‘ฅ-intercepts. Doing so wouldโ€™ve given us ๐‘ฅ equals one and ๐‘ฅ equals 3.5, which once again matches function (A). So the graph that represents the equation given is graph (A).

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