We can say at month 0 she was 43 inches tall.
And at month 18 she was 52 inches tall.
We can say that x is the number of months and y is her height.
Rise/run
y2-y1/x2-x1
52-43/18-0
9/18
1/2
So the slope is 1/2 (which is what you are looking for.) and if you were to write an equation it would be y=1/2x+43
Brainliest my answer if it helps you out?
Answer:
3people/minute
Step-by-step explanation:
75/25=3
Steps:
1) determine the domain
2) determine the extreme limits of the function
3) determine critical points (where the derivative is zero)
4) determine the intercepts with the axis
5) do a table
6) put the data on a system of coordinates
7) graph: join the points with the best smooth curve
Solution:
1) domain
The logarithmic function is defined for positive real numbers, then you need to state x - 3 > 0
=> x > 3 <-------- domain
2) extreme limits of the function
Limit log (x - 3) when x → ∞ = ∞
Limit log (x - 3) when x → 3+ = - ∞ => the line x = 3 is a vertical asymptote
3) critical points
dy / dx = 0 => 1 / x - 3 which is never true, so there are not critical points (not relative maxima or minima)
4) determine the intercepts with the axis
x-intercept: y = 0 => log (x - 3) = 0 => x - 3 = 1 => x = 4
y-intercept: The function never intercepts the y-axis because x cannot not be 0.
5) do a table
x y = log (x - 3)
limit x → 3+ - ∞
3.000000001 log (3.000000001 -3) = -9
3.0001 log (3.0001 - 3) = - 4
3.1 log (3.1 - 3) = - 1
4 log (4 - 3) = 0
13 log (13 - 3) = 1
103 log (103 - 3) = 10
lim x → ∞ ∞
Now, with all that information you can graph the function: put the data on the coordinate system and join the points with a smooth curve.
This gives you three simultaneous equations:
6 = a + c
7 = 4a + c
1 = c
<u>c = 1
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If c =1,
6 = a + 1
<u>a = 5
</u><u /><u />
This doesn't work in the second equation, so the quadratic that goes through these points is not in the form y = ax^2 + bx + c
Was there supposed to be a b in the equation?
-10/8 and + 15/8. Those should work.
