2 answers:
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Make it into slope intercept form which is y=mx+b
subtract x ffrom both sides
2y=-5-x
divide by 2
y=-1/2x-5/2
y=mx+b
m=slope
b=y intercept or where the line passes through the y-axis
now one way is to graph points by subsituting
if x is 0 then y=-5//2
if x=1 then y=-3
if x=-5 then y=0
another way is to find out where the line passes though the y and x axis or the y and x intecepts respectively
the y intercept is when x=0
the x intecept is when y=0
so set each to zero and solve
if x=0
y=-1/2(0)-5/2
y=-5/2
the line passes throught the y axis at -5/2, plot the point (0,-5/2) on the y axis
then set y to zero
0=-1/2x-5/2
add 5/2 to both sides
5/2=-1/2x
multiply -1
-5/2=1/2x
multiply by 2
-5=x
therefor the line passes through tthe x axis at (-5,0)
then draw the line between the 2 points
here is the graph
subtract x ffrom both sides
2y=-5-x
divide by 2
y=-1/2x-5/2
y=mx+b
m=slope
b=y intercept or where the line passes through the y-axis
now one way is to graph points by subsituting
if x is 0 then y=-5//2
if x=1 then y=-3
if x=-5 then y=0
another way is to find out where the line passes though the y and x axis or the y and x intecepts respectively
the y intercept is when x=0
the x intecept is when y=0
so set each to zero and solve
if x=0
y=-1/2(0)-5/2
y=-5/2
the line passes throught the y axis at -5/2, plot the point (0,-5/2) on the y axis
then set y to zero
0=-1/2x-5/2
add 5/2 to both sides
5/2=-1/2x
multiply -1
-5/2=1/2x
multiply by 2
-5=x
therefor the line passes through tthe x axis at (-5,0)
then draw the line between the 2 points
here is the graph
Answer:
- a ≈ 615.272
- b ≈ 1.61875
Step-by-step explanation:
My graphing calculator can solve this problem two ways. It can do a least-squares fit of the exponential data, or it can do a least-squares fit of the logarithm of the data. The answers come out different.
In "log mode", the parameters of the exponential equation are computed as ...
a ≈ 615.272
b ≈ 1.61875
__
With log mode turned off, the "a" parameter is lower (≈553.071), and the "b" parameter is higher (≈1.65573). This mode gives a better fit to the larger data values, at the expense of a poorer fit to the lower data values, as you might expect.
