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course phy 242
1)y= 3x-4 x y
-2 -10
-1 -7
0 -4
1 -1
2 2
The y-intercept of the graph is -4 due to the given linear equation y = mx+b, where b is y-intercept. The point where the line goes through the x-axis is determined through setting y equal to 0. 0=3x-4 4=3x x= 4/3
Therefore, the graph crosses the x-axis at the point (4/3,0).
The drawn graph confirms the information given about it.
2) In this problem “steepness” can be referring to the slope of the graph. Remembering the equation y = mx+b , m is the slope so therefore, 3 is the slope of the graph. The slope of a linear graph does not change, so therefore, the steepness of the graph does not change.
3) Taking the points (-2,-10) and (0,-4)
Rise/run = (-10-(-4))/(-2-0) = (-10+4)/(-2)= -6/-2= 3
The slope of the graph is 3 when calculated from using two point on the line.
4)x y
0 0
1 1
2 4
3 9
The graph is increasing in the direction of positive x values.
Yes the steepness of the graph changes. It gets steeper as the values of x increases. One could look at the first derivative which is y’=2x , which is positive. Implying positive change.
The graph is increasing at an increasing rate because with each increasing x value the y value gets greater and greater. One could also look at the second derivative which is y’’=2 , which is positive which demonstrates that the function is increasing at an increasing rate.
5) x y
-3 9
-2 4
-1 1
0 0
The graph is decreasing through to the decreasing values of the y. The rate is also decreasing.
Yes the steepness of the graph changes. It because less steeper as the x value increases.
It could be said that the graph is decreasing at a decreasing rate which is demonstrated through the second derivative y’’= 2. Since we know the slope is negative through the points a positive second derivative means that the graph is decreasing in a decreasing rate.
6)x y
0 0
1 1
2 sqrt(2)
3 sqrt(3)
The graph is increasing.
The first derivative of the function is y’= 1/(2*sqrt(x)),therefore as the x value increases the slope decreases and so does the steepness.
The second derivative of the graph is y’’= - 1 /(4x*sqrt(x)). This demonstrates that the second derivative is negative and so it can be said that the graph is increasing at a decreasing rate.
7)x y
0 5
1 5/2
2 5/4
3 5/8
The graph is decreasing due to the decreasing values of y.
The first derivative of the function is y’=( -5*log(2))/(2^x). This demonstrates that as the graph decreases, while getting less steeper.
The second derivative is y’’= 5 *2^(-x) *log^2(2). This positive second derivative demonstrates that the graph is concave up. Therefore , the graph is decreasing at a decreasing rate.
8) The graph of y versus t would be an increasing graph, because as t increases so does the distance of the car from me. Since the car is moving faster and faster, the graph would increase in an increasing rate.
9)x y x y
-3 16 1 0
-2 9 2 1
-1 4 3 4
0 1
From x=-3 to x=1 the graph of y=(x-1)^2 is decreasing at a decreasing rate.
From x=1 to x=3 the graph of y=(x-1)^2 is increasing at an increasing rate.
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Very good, and still no problem, but see my previous notes about submitting the entire document. This is going to become important.
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