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What is the distance, in units, between the y-intercept of f(x) = x2 + 7x - 18 and the y-intercept of the linear function that passes through
the points shown in the table below?
х
g(x)
-5
2
10
11
25
20
60
41


What Is The Distance In Units Between The Yintercept Of Fx X2 7x 18 And The Yintercept Of The Linear Function That Passes Through The Points Shown In The Table class=

Sagot :

Answer:

y = 0.6x - 5

Step-by-step explanation:

The distance between the y-intercept of f(x) = x^2 + 7x - 18 and the y-intercept of the table is 23 units

How to determine the distance?

The function f(x) is given as:

f(x) = x^2 + 7x - 18

The y-intercept is when x = 0.

So, we have:

f(0) = 0^2 + 7(0) - 18

Evaluate

f(0) = -18

For the table, start by calculating the slope (m)

m = 11 - 2/10 + 5

m = 0.6

The y-intercept is then calculated using:

b = y - mx

So, we have:

b = 11 - 0.6 * 10

Evaluate

b = 5

This means that:

y-intercept = 5

The difference between both values is:

Difference = |-18- 5|

Evaluate

Difference = 23

Hence, the distance between the y-intercept of f(x) = x^2 + 7x - 18 and the y-intercept of the table is 23 units

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