QuestionDetermine if the function defined by the points (4, 96), (5, 150), (6, 216), (7, 294), (8, 384) is linear, quadratic, or exponential.

Studdy Solution

STEP 1

Assumptions1. We have a set of data points $(x, y)$ . We need to determine if the relationship between $x$ and $y$ is linear, quadratic, or exponential

STEP 2

First, let's check if the function is linear. A function is linear if the difference between consecutive $y$ values (also known as the change in $y$ , or $\Delta y$ ) is constant. $\Delta y = y_{n+1} - y_n$

STEP 3

Calculate the difference between consecutive $y$ values.
$\Delta y1 =150 -96 =54$ $\Delta y2 =216 -150 =66$ $\Delta y3 =294 -216 =78$ $\Delta y =384 -294 =90$

STEP 4

Since the differences between consecutive $y$ values are not constant, the function is not linear.

STEP 5

Next, let's check if the function is quadratic. A function is quadratic if the difference between consecutive $\Delta y$ values (also known as the second difference, or $\Delta^2 y$ ) is constant.
$\Delta^2 y = \Delta y_{n+1} - \Delta y_n$

STEP 6

Calculate the second difference.
$\Delta^2 y1 =66 -54 =12$ $\Delta^2 y2 =78 -66 =12$ $\Delta^2 y3 =90 -78 =12$

STEP 7

Since the second differences are constant, the function is quadratic.
Therefore, the function represented by the given data points is quadratic.

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QuestionDetermine if the function defined by the points (4, 96), (5, 150), (6, 216), (7, 294), (8, 384) is linear, quadratic, or exponential.

Studdy Solution

STEP 1

Assumptions1. We have a set of data points $(x, y)$ . We need to determine if the relationship between $x$ and $y$ is linear, quadratic, or exponential

STEP 2

First, let's check if the function is linear. A function is linear if the difference between consecutive $y$ values (also known as the change in $y$ , or $\Delta y$ ) is constant. $\Delta y = y_{n+1} - y_n$

STEP 3

Calculate the difference between consecutive $y$ values.
$\Delta y1 =150 -96 =54$ $\Delta y2 =216 -150 =66$ $\Delta y3 =294 -216 =78$ $\Delta y =384 -294 =90$

STEP 4

Since the differences between consecutive $y$ values are not constant, the function is not linear.

STEP 5

Next, let's check if the function is quadratic. A function is quadratic if the difference between consecutive $\Delta y$ values (also known as the second difference, or $\Delta^2 y$ ) is constant.
$\Delta^2 y = \Delta y_{n+1} - \Delta y_n$

STEP 6

Calculate the second difference.
$\Delta^2 y1 =66 -54 =12$ $\Delta^2 y2 =78 -66 =12$ $\Delta^2 y3 =90 -78 =12$

STEP 7

Since the second differences are constant, the function is quadratic.
Therefore, the function represented by the given data points is quadratic.

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QuestionDetermine if the function defined by the points (4, 96), (5, 150), (6, 216), (7, 294), (8, 384) is linear, quadratic, or exponential.

Studdy Solution

STEP 1

Assumptions1. We have a set of data points (x,y)(x, y)(x,y). We need to determine if the relationship between xxx and yyy is linear, quadratic, or exponential

STEP 2

First, let's check if the function is linear. A function is linear if the difference between consecutive yyy values (also known as the change in yyy, or Δy\Delta yΔy) is constant.Δy=yn+1−yn\Delta y = y_{n+1} - y_nΔy=yn+1​−yn​

STEP 3

Calculate the difference between consecutive yyy values.Δy1=150−96=54\Delta y1 =150 -96 =54Δy1=150−96=54Δy2=216−150=66\Delta y2 =216 -150 =66Δy2=216−150=66Δy3=294−216=78\Delta y3 =294 -216 =78Δy3=294−216=78Δy=384−294=90\Delta y =384 -294 =90Δy=384−294=90

STEP 4

Since the differences between consecutive yyy values are not constant, the function is not linear.

STEP 5

Next, let's check if the function is quadratic. A function is quadratic if the difference between consecutive Δy\Delta yΔy values (also known as the second difference, or Δ2y\Delta^2 yΔ2y) is constant.Δ2y=Δyn+1−Δyn\Delta^2 y = \Delta y_{n+1} - \Delta y_nΔ2y=Δyn+1​−Δyn​

STEP 6

Calculate the second difference.Δ2y1=66−54=12\Delta^2 y1 =66 -54 =12Δ2y1=66−54=12Δ2y2=78−66=12\Delta^2 y2 =78 -66 =12Δ2y2=78−66=12Δ2y3=90−78=12\Delta^2 y3 =90 -78 =12Δ2y3=90−78=12

STEP 7

Since the second differences are constant, the function is quadratic.Therefore, the function represented by the given data points is quadratic.

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Assumptions1. We have a set of data points $(x, y)$ . We need to determine if the relationship between $x$ and $y$ is linear, quadratic, or exponential

First, let's check if the function is linear. A function is linear if the difference between consecutive $y$ values (also known as the change in $y$ , or $\Delta y$ ) is constant. $\Delta y = y_{n+1} - y_n$

Calculate the difference between consecutive $y$ values.
$\Delta y1 =150 -96 =54$ $\Delta y2 =216 -150 =66$ $\Delta y3 =294 -216 =78$ $\Delta y =384 -294 =90$

Since the differences between consecutive $y$ values are not constant, the function is not linear.

Next, let's check if the function is quadratic. A function is quadratic if the difference between consecutive $\Delta y$ values (also known as the second difference, or $\Delta^2 y$ ) is constant.
$\Delta^2 y = \Delta y_{n+1} - \Delta y_n$

Calculate the second difference.
$\Delta^2 y1 =66 -54 =12$ $\Delta^2 y2 =78 -66 =12$ $\Delta^2 y3 =90 -78 =12$

Since the second differences are constant, the function is quadratic.
Therefore, the function represented by the given data points is quadratic.