Answer:
27abc
you multiply the coefficient and write all the variables after
Y=mx+b
m=slope
b=yintercept
slope=(y1-y2)/(x1-x2)
(x,y)
x1=6
y1=-4
x2=-6
y2=2
(-4-2)/(6-(-6))=-6/12=-1/2
y=-1/2x+b
subsitute one point
(6,-4)
x=6
y=-4
-4=1/2(6)+b
-4=3+b
subtract 3
-7=b
y=-1/2x-7 is answer
I think you forgot to mention Susan's current age if not:
x+5
x-4
6x
The generic equation for a linear function can be expressed in the slope intercept form f(x) = mx + b, where m is the slope and b is the y intercept. For this problem we can first find the equation of the line. Then we substitute x = 7 to get the f(x) value, which is n at the point x = 7.
To find the equation of the linear function we first find the slope. Slope is defined as the change in f(x) divided by the change in x. As we are given a linear function, the slope at every point is the same. We can pick any two points known to find the slope.
Let's pick (3, 7) and (9, 16). The slope m is m = (16-7)/(9-3) = 9/6 = 3/2.
Now that we have the slope, we can plug in the slope and one of the points to find b. Let's use the point (3, 7).
f(x) = mx + b
7 = (1/2)(3) + b
b = 11/2
Now we can write the equation
f(x) = (1/2)x + 11/2
Plugging in x = 7 we find that f(7) = 9. n = 9
The tangent line to a curve is the one that coincides with the curve at a point and with the same derivative, that is, the same degree of variation.
We have then:
y = 5x-x²
Deriving:
y '= 5-2x
In point (1, 4)
The slope is:
y (1) '= 5-2 * (1)
y (1) '= 3
The equation of the line will be:
y-f (a) = f '(a) (x-a)
We have then:
y-4 = 3 (x-1)
Rewriting:
y = 3x-3 + 4
y = 3x + 1
Answer:
the tangent line to the parabola at the point (1, 4) is
y = 3x + 1
the slope m is
m = 3
