15.
Do the data in the table represent a linear function? If so, write a rule for the function.
x –3 –2 –1 0 1
y 1 –2 –5 –8 –11
(1 point)
yes; y = –3x – 8
yes; y = 1/3x – 8
yes; y = 1/3x + 8
yes; y = 3x + 8
16.
Write a quadratic rule for the data in the table.
x –1 0 1 2 3
y 4 5 4 1 –4
(1 point)
y = –2x^2 + 5
y = –x^2 + 5
y = x^2 – 5
y = x^2 + 5
Can you please also explain how to solve this?
a. for each decrease of 3 in y,an increase of 1 in x.
y=mx+b=-3/1 m +b
and b..
1=-3(-3)+b
b=8
b. y=ax^2+bx+c
put in three different data points, solve for a, b, c.
for the quadratic, note that
f(-1) = f(1) = 4
That means the vertex lies on the line x=0.
y = a(x-0)^2 + c
or,
y = ax^2+5
Now use one other point to get a:
1 = 4a+5
a = -1
y = -x^2+5
To determine if the data in the table represents a linear function, we need to check if there is a constant rate of change between the x and y values.
In the first table, let's calculate the rate of change between each pair of x and y values:
(0-(-1))/(5-4) = 1/1 = 1
(-1-(-2))/(-5-(-2)) = 1/3
(-2-(-3))/(-8-(-5)) = 1/3
(-3-(-4))/(-11-(-8)) = 1/3
Since the rate of change is constant at 1/3, we can conclude that the data in the first table represents a linear function.
To find the equation for a linear function, we can use the slope-intercept form, which is y = mx + b. In this form, m represents the slope and b represents the y-intercept.
The slope, m, is equal to the rate of change we calculated earlier, which is 1/3.
To find the y-intercept, we can substitute any of the x and y values from the table into the equation and solve for b. Let's choose the first pair of values (-3, 1):
1 = (1/3)(-3) + b
1 = -1 + b
b = 2
Therefore, the equation for the linear function is y = (1/3)x + 2.
For the second question,
To determine if the data in the table represents a quadratic function, we need to check if there is a constant second difference between the y values.
In the second table, let's calculate the difference between each pair of y values:
5 - 4 = 1
4 - 5 = -1
1 - 4 = -3
-4 - 1 = -5
Since the second difference is constant at -2, we can conclude that the data in the second table represents a quadratic function.
To find the equation for a quadratic function, we can use the form y = ax^2 + bx + c. In this form, a, b, and c represent coefficients that we need to determine.
To find the coefficients, we can substitute any three of the x and y values into the equation and solve for a, b, and c. Let's choose the first three pairs of values (-1, 4), (0, 5), and (1, 4):
4 = a(-1)^2 + b(-1) + c
5 = a(0)^2 + b(0) + c
4 = a(1)^2 + b(1) + c
These equations simplify to:
4a - b + c = 4 (equation 1)
c = 5
a + b + c = 4 (equation 2)
By substituting c = 5 into equation 1, we get:
4a - b + 5 = 4
By substituting c = 5 into equation 2, we get:
a + b + 5 = 4
Simplifying further, we have:
4a - b = -1 (equation 3)
a + b = -1 (equation 4)
Adding equations 3 and 4 together, we have:
5a = -2
Dividing both sides by 5, we find:
a = -2/5
Substituting a = -2/5 in equation 4, we find:
(-2/5) + b = -1
b = -1 + 2/5
b = -3/5
Therefore, the equation for the quadratic function is y = (-2/5)x^2 - (3/5)x + 5.