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Question Video: Determining Which Graph Is That of the Square Root of a Quadratic Function Mathematics

Which of the following is the graph of the function 𝑓(π‘₯) = √(π‘₯Β² + π‘₯ βˆ’ 6)? [A] Graph A [B] Graph B [C] Graph C [D] Graph D [E] Graph E

10:19

Video Transcript

Which of the following is the graph of the function 𝑓 of π‘₯ is equal to the square root of π‘₯ squared plus π‘₯ minus six? Is it graph (A), graph (B), (C), (D), or graph (E)?

To identify the correct graph, the first thing we note is that the function 𝑓 of π‘₯ is a radical function, that is, a function containing a square root. And when considering the graph of a radical function, it’s a good idea to begin by thinking about the domain and range of the function. Remember, the domain is the set of π‘₯-values or input values the function can take, and the range is the set of possible output or 𝑦-values. We’re going to begin by considering the range. We see in fact that 𝑓 of π‘₯ consists of a positive square root of a quadratic function. And because the square root is greater than or equal to zero, 𝑓 of π‘₯ is never negative.

Now, we know that any quadratic function of the form π‘Žπ‘₯ squared plus 𝑏π‘₯ plus 𝑐 whose leading coefficient π‘Ž is positive is U-shaped. This means that as π‘₯ tends to either positive or negative ∞, the quadratic function approaches positive ∞. We also know that the positive square root function approaches ∞ as π‘₯ tends to ∞. So the positive square root of a quadratic with positive leading coefficient will also go to ∞.

Now looking at the given graphs, they all hit the π‘₯-axis at the two points π‘₯ is negative three and π‘₯ is equal to two. And checking these values in our function 𝑓, we find 𝑓 of negative three is equal to zero and 𝑓 of two is also equal to zero. Hence, the range of our function 𝑓 includes zero and is the left-closed, right-open interval from zero to ∞. So actually at this point, we can eliminate the four graphs (B), (C), (D), and (E). We can do this because the parts of the graphs (B), (C), (D), and (E) we’ve highlighted are below the π‘₯-axis and therefore negative. Hence, none of these four graphs can be our function, since 𝑓 of π‘₯ must be greater than or equal to zero.

This leaves us with graph (A), since this is the only one of the five with no negative values for 𝑓 of π‘₯. Now, we might at this point think our work is done. But how can we confirm that graph (A) is actually the graph of our function 𝑓? Well, there are a few things we can consider, but let’s start by looking at the domain of 𝑓. Remember, the domain is the possible π‘₯- or input values a function can take. Since 𝑓 is a real-valued function, this means that the argument of the square root or radical cannot be negative. That is, π‘₯ squared plus π‘₯ minus six must be greater than or equal to zero. Otherwise, we would have the square root of a negative number, a complex number, which is not possible since 𝑓 is a real-valued function.

To solve this inequality for π‘₯, let’s first solve the equation π‘₯ squared plus π‘₯ minus six is equal to zero. We can do this using the quadratic formula or any other method, and we find the solutions π‘₯ is equal to negative three and π‘₯ is positive two. We see that this matches with graph (A), where the function meets the π‘₯-axis at π‘₯ equals negative three and at π‘₯ equals two. Our next step is to check whether the function 𝑓 of π‘₯ exists within each interval of π‘₯-values, that is, for π‘₯ less than negative three, π‘₯ between negative three and positive two, and for π‘₯ greater than two. If our function matches graph (A), we would expect to find that 𝑓 is positive for π‘₯ less than negative three and greater than two and that it doesn’t exist for π‘₯ between these two values.

Now, we noted previously that if a quadratic function has positive leading coefficient, its graph is U-shaped. We also know that if a quadratic function has two real roots, then it is the same sign for π‘₯-values to the left of the lower-valued root and to the right of the higher-valued root and the opposite sign between the two roots. In our case with positive leading coefficient, the quadratic is positive to the right of π‘₯ equals two and to the left of π‘₯ equals negative three and negative between the two roots. This means that our square root function must also be positive for π‘₯ greater than two and less than negative three. Hence, 𝑓 is positive and exists in these two regions.

Considering π‘₯-values between negative three and positive two, however, since the quadratic is negative there and the square root of a negative number is not real valued, 𝑓 of π‘₯ does not exist for π‘₯ greater than negative three and less than positive two. And we see that again this matches with graph (A).

So far so good with our function 𝑓 and graph (A). The domain of 𝑓 is the same as that of graph (A). That’s π‘₯ less than or equal to negative three or π‘₯ greater than or equal to positive two. And remember, our ranges also match.

Something else we can consider to determine whether graph (A) matches our function 𝑓 of π‘₯ is the derivative or slope of 𝑓 of π‘₯. That’s 𝑓 prime or d𝑓 by dπ‘₯. We see in graph (A) that as π‘₯ approaches negative three from the left, the graph is a decreasing function. This means that the slope or gradient is negative. And if 𝑓 of π‘₯ matches this graph, then 𝑓 prime should be negative in this region. Similarly, as π‘₯ gets larger to the right of positive two, graph (A) is an increasing function, so its slope is positive there. So we want 𝑓 prime of π‘₯ to also be positive for π‘₯-values in this region. Let’s test this by finding 𝑓 prime and trying some appropriate values in the result.

Now, recall that a square root or radical of a number or expression is actually that number or expression to the power of one-half. So we can express 𝑓 in this form. Now using the result that d by dπ‘₯ of a function 𝑒 raised to the power 𝑛, where 𝑒 is a function of π‘₯, is equal to 𝑛 multiplied by 𝑒 raised to the power 𝑛 minus one times d𝑒 by dπ‘₯, with 𝑒 equals π‘₯ squared plus π‘₯ minus six, 𝑛, the exponent equal to one over two, we have d𝑒 by dπ‘₯ equal to two π‘₯ plus one. This means that 𝑓 prime is one-half multiplied by π‘₯ squared plus π‘₯ minus six, that’s 𝑒, all raised to the power negative one-half, which is 𝑛 minus one, multiplied by d𝑒 by dπ‘₯, which is two π‘₯ plus one.

Recalling then that an expression raised to a negative power is one over the expression to that positive power, we have 𝑓 prime of π‘₯ is equal to two π‘₯ plus one over two times π‘₯ squared plus π‘₯ minus six raised to the power of one-half. And we can convert our exponent of one-half back into a square root as shown.

We know that our function 𝑓 is continuous to the left of π‘₯ equals negative three and to the right of π‘₯ equals two, since it’s the positive square root of a continuous function in these regions. And we can now find 𝑓 prime of π‘₯ for π‘₯-values within these two regions to determine whether 𝑓 is increasing or decreasing there. We can do this because if 𝑓 prime of π‘₯ is positive for π‘₯ in the open interval π‘Ž to 𝑏, then we know that 𝑓 is increasing on that interval. And if 𝑓 prime of π‘₯ is less than zero for π‘₯ in the open interval π‘Ž to 𝑏, then we know that 𝑓 is decreasing on that interval.

Let’s take one π‘₯-value in each interval, say, π‘₯ equal to negative five and π‘₯ equals positive four. First, with π‘₯ equals negative five, we have 𝑓 prime of negative five is equal to two times negative five plus one over two times the square root of negative five squared plus negative five minus six, that is, negative 10 plus one over two times the square root of 14. And this evaluates to negative nine over two times the square root of 14. This is negative, so as we require, for 𝑓 to match graph (A), the slope of 𝑓 of π‘₯ for this value of π‘₯ is negative. Similarly, 𝑓 prime of four is positive nine over two times the square root of 14, which is greater than zero. Hence, the slope of 𝑓 of π‘₯ at π‘₯ equals four is positive, which again is as we require and matches the behavior of graph (A).

By finding the derivative of 𝑓 of π‘₯ then and trying some points in the two regions in which the function exists, we find that the sign of the slope of 𝑓 for points in those regions matches that of graph (A).

One final thing we can consider is whether or not 𝑓 of π‘₯ has any critical points. A function 𝑓 is said to have a critical point, where 𝑓 prime of π‘₯ is either equal to zero or does not exist. If 𝑓 prime of π‘₯ is equal to zero, then there may be a turning point at π‘₯. And if 𝑓 is continuous, on any closed interval π‘Ž, 𝑏, the absolute extrema occur at the critical points of 𝑓 or at the endpoints of the closed interval. We see that graph (A) does not have any turning points, so if it is the graph of 𝑓 of π‘₯, then neither should 𝑓 of π‘₯.

For our function 𝑓, the only way for 𝑓 prime of π‘₯ to equal zero is if the numerator equals zero, that is, if two π‘₯ plus one is equal to zero. Solving this, we find π‘₯ equal to negative one-half. However, our function 𝑓 does not exist for π‘₯ equal to negative one-half. So there is no value of π‘₯ where a turning point might occur. We can also try to find any values of π‘₯ for which 𝑓 prime does not exist. In our case, these may occur at values of π‘₯ for which the denominator is equal to zero. We already know the solutions to this however. That’s where π‘₯ equals negative three and π‘₯ equals positive two. This again matches with graph (A) since as we see, π‘₯ equal to negative three and π‘₯ is positive two are critical points of 𝑓 and are the endpoints of the intervals of the domain.

Our function 𝑓 and graph (A) therefore have matching range, domain, and critical points, where the endpoints are at π‘₯ equals negative three and π‘₯ equals two and where π‘₯ is negative one-half is outside of the domain of 𝑓. We know also that both functions, 𝑓 and graph (A), are decreasing on the interval negative ∞ to negative three and that both 𝑓 and graph (A) are increasing on the interval from π‘₯ equals two to positive ∞. Hence, we can say that the graph corresponding to the function 𝑓 of π‘₯ is equal to the positive square root of π‘₯ squared plus π‘₯ minus six is graph (A).

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