Which linear equation includes the points (6,0) and (0, – 4)?
1 answer:
Answer:
y= (2/3)x - 4
Step-by-step explanation:
To answer this you'll want to use the slope intercept form of a line:
y= mx + b
So to solve essentially you just have to find the values for m and b
- m represents the slope of the line
- b is the y-intercept (the point where the line crosses the y-axis)
You can figure that b = -4 because the line will cross the y-axis when x = 0, and you are given the point on the line (0,-4), where x is equal to 0.
Now to find the slope of the line, since you are given two points on the line you will want to use the slope equation:
slope =
So you can assign each of the two points a number (1 or 2) where its values are equal to (x1, y1) or (x2, y2). It does not matter which point gets which number.
In this example:
We will assign the point (6,0) to be (x1, y1),
and we will assign the point (0, -4) to be (x2, y2)
So we end up with the following:
m=
m=
m=
Simplify the fraction:
m =
From there you just substitute in for m and b into the slope intercept equation
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Answer:
a) , b)
, c) Concave upward:
. No intervals being concave downward.
Step-by-step explanation:
a) Tangent line is vertical if slope is undefined, whose points are associated with discontinuities of the function. Rational functions have discontinuities associated with the denominator:
By factorizing each component, the function is re-arranged as:
There is no avoidable discontinuities. The only point where tangent line is vertical is .
b) Tangent line is horizontal if slope is equal to zero, whose point is associated to the points that makes function equal to zero. Rational functions have horizontal tangent lines if numerator is equal to zero and denominator is different to zero. Hence, the only point where tangent line is horizontal is .
c) An interval is concave upward if exist an absolute minimum inside, which can be found by the help of the First and Second Derivative Tests.
The only critical point is:
(which coincides with the result of point b)
The value associated with the critical point is:
Which means that critical point is an absolute minimum, and, consequently, the interval that is concave upward is . There is no absolute maximums and, therefore, there is no interval that is concave downward.