Video: US-SAT04S3-Q13-749108140360

How do the graphs of 𝑦 = 2π‘₯Β² + 5 and of 𝑦 = π‘₯Β² + 5 differ?

04:47

Video Transcript

How do the graphs of 𝑦 equals two π‘₯ squared plus five and of 𝑦 equals π‘₯ squared plus five differ?

We could answer this question using our knowledge of transformations of graphs. Alternatively, we could draw them both on coordinate axes. Let’s, firstly consider the graph of 𝑦 equals π‘₯ squared plus five. Any quadratic graph where the coefficient of π‘₯ squared is positive will be U shaped. A quadratic graph is one where the highest power of π‘₯ is two. The graph will have a 𝑦-intercept of five, as this is the constant. As there is no π‘₯ term, this will also be the minimum point of the graph. We can work out some other coordinates on the graph by substituting in some negative and positive π‘₯-values.

Negative two squared is equal to four as multiplying a negative number by a negative number gives us a positive answer. Adding five to this gives us nine. Therefore, when π‘₯ is equal to negative two, 𝑦 is equal to nine. Substituting in negative one, zero, one, and two into the equation, 𝑦 equals π‘₯ squared plus five, gives us values of six, five, six, and nine, respectively. We can then plot these five points in the π‘₯𝑦-plane. We end up with a U-shaped graph with minimum points zero, five as previously mentioned. We can now repeat the same process for the equation 𝑦 equals two π‘₯ squared plus five.

This, too, is a quadratic graph, therefore, once again will be U-shaped. We need to substitute in the same values of π‘₯ to compare the two graphs. Substituting in π‘₯ equals negative two gives us 𝑦 is equal to two multiplied by negative two squared plus five. In order to work out our value of 𝑦, we need to use our older of operations. Our first step is to do the exponents or powers. Negative two squared, as previously mentioned, is equal to four. So we’re left with two multiplied by four plus five. Next, we need to multiply two and four. This gives us eight. Finally, eight plus five is equal to 13.

This means that when π‘₯ is equal to negative two the value for 𝑦 is 13. Substituting in π‘₯ equals negative one, zero, one, and two gives us 𝑦 values of seven, five, seven, and 13, respectively. We can once again plot these five points on the π‘₯𝑦-plane. Once again, we have a U-shaped parabola with minimum point zero, five. We were asked in this question to describe how the two graphs differ. We can see that the minimum points of both graphs is the same. However, the 𝑦-values at the other corresponding π‘₯ points are different. The values in the equation 𝑦 equals two π‘₯ squared plus five are greater than 𝑦 equals π‘₯ squared plus five. For example, 13 is greater than nine and seven is greater than six. This means that the graph has a steeper slope.

We can, therefore, conclude that the graph of 𝑦 equals two π‘₯ squared plus five is narrower or steeper than the graph of 𝑦 equals π‘₯ squared plus five. As the π‘₯ squared term in our equation has been multiplied by two, we can go one stage further here using our knowledge of transformations of graphs. We know that the graph of 𝑓 of two π‘₯ is a horizontal stretch of 𝑓 of π‘₯ by a factor of one-half. This means that the graph of 𝑓 of two π‘₯ will be narrower or steeper than the graph of 𝑓 of π‘₯. In this example, this means that 𝑦 equals two π‘₯ squared plus five is narrower than 𝑦 equals π‘₯ squared plus five.

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