Question Find the equation that best fits the data set {(-4,-4.8),(-3,-8.2),(-2,-9.1),(-1,-8.1),(0,-4.7),(1,0.3)}. (a) $y=1.1x+4.2$ (b) $y=1.1x^2+4.2x+4.9$ (c) $y=1.1x^2+4.2x-4.9$ (d) $y=1.1x-4.2$

Studdy Solution

STEP 1

Assumptions
1. We have a data set of points in the form $(x, y)$ .
2. We need to determine which equation best represents the given data set.
3. The options provided are linear and quadratic equations.

STEP 2

We will first test the linear option (巨) $y = 1.1x + 4.2$ by plugging in the x-values from the data set and checking if the corresponding y-values match.

STEP 3

Test the point $(-4, -4.8)$ in the equation $y = 1.1x + 4.2$ .
$y = 1.1(-4) + 4.2$

STEP 4

Calculate the y-value for $x = -4$ .
$y = -4.4 + 4.2 = -0.2$

STEP 5

Compare the calculated y-value with the given y-value in the data set for $x = -4$ .
The calculated y-value $-0.2$ does not match the given y-value $-4.8$ , so option (巨) is not the correct equation.

STEP 6

Next, we will test the quadratic option (G) $y = 1.1x^2 + 4.2x + 4.9$ using the same point $(-4, -4.8)$ .
$y = 1.1(-4)^2 + 4.2(-4) + 4.9$

STEP 7

Calculate the y-value for $x = -4$ .
$y = 1.1(16) - 16.8 + 4.9$

STEP 8

Simplify the calculation.
$y = 17.6 - 16.8 + 4.9$

STEP 9

Finish the calculation to find the y-value.
$y = 0.8 + 4.9 = 5.7$

STEP 10

Compare the calculated y-value with the given y-value in the data set for $x = -4$ .
The calculated y-value $5.7$ does not match the given y-value $-4.8$ , so option (G) is not the correct equation.

STEP 11

Now, we will test the quadratic option (円) $y = 1.1x^2 + 4.2x - 4.9$ using the same point $(-4, -4.8)$ .
$y = 1.1(-4)^2 + 4.2(-4) - 4.9$

STEP 12

Calculate the y-value for $x = -4$ .
$y = 1.1(16) - 16.8 - 4.9$

STEP 13

Simplify the calculation.
$y = 17.6 - 16.8 - 4.9$

STEP 14

Finish the calculation to find the y-value.
$y = 0.8 - 4.9 = -4.1$

STEP 15

Compare the calculated y-value with the given y-value in the data set for $x = -4$ .
The calculated y-value $-4.1$ is close to the given y-value $-4.8$ , but not exact. We should test another point from the data set to confirm if this equation is correct.

STEP 16

Test the point $(0, -4.7)$ in the equation $y = 1.1x^2 + 4.2x - 4.9$ .
$y = 1.1(0)^2 + 4.2(0) - 4.9$

STEP 17

Calculate the y-value for $x = 0$ .
$y = -4.9$

STEP 18

Compare the calculated y-value with the given y-value in the data set for $x = 0$ .
The calculated y-value $-4.9$ is close to the given y-value $-4.7$ , but not exact. We should test another point to confirm if this equation is correct.

STEP 19

Test the point $(1, 0.3)$ in the equation $y = 1.1x^2 + 4.2x - 4.9$ .
$y = 1.1(1)^2 + 4.2(1) - 4.9$

STEP 20

Calculate the y-value for $x = 1$ .
$y = 1.1 + 4.2 - 4.9$

STEP 21

Finish the calculation to find the y-value.
$y = 5.3 - 4.9 = 0.4$

STEP 22

Compare the calculated y-value with the given y-value in the data set for $x = 1$ .
The calculated y-value $0.4$ is very close to the given y-value $0.3$ . This suggests that option (円) $y = 1.1x^2 + 4.2x - 4.9$ is likely the correct equation. However, to be certain, we should test all points or at least one more point from the data set.

STEP 23

Test the point $(-2, -9.1)$ in the equation $y = 1.1x^2 + 4.2x - 4.9$ .
$y = 1.1(-2)^2 + 4.2(-2) - 4.9$

STEP 24

Calculate the y-value for $x = -2$ .
$y = 1.1(4) - 8.4 - 4.9$

STEP 25

Finish the calculation to find the y-value.
$y = 4.4 - 8.4 - 4.9 = -8.9$

STEP 26

Compare the calculated y-value with the given y-value in the data set for $x = -2$ .
The calculated y-value $-8.9$ is very close to the given y-value $-9.1$ . This further supports that option (円) $y = 1.1x^2 + 4.2x - 4.9$ is the correct equation.

STEP 27

Since the quadratic equation (円) $y = 1.1x^2 + 4.2x - 4.9$ has provided y-values close to the given y-values for multiple points in the data set, we can conclude that it is the best representation of the data set.
The correct equation that best represents the data set is:
$y = 1.1x^2 + 4.2x - 4.9$

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