Create a graph that has the following intercepts. make sure to label each of the intercepts on
the graph, as well as the axis on your graph.
y=-3, x = 5, y = 3

  see below for the graph

Step-by-step explanation:

The desired graph has two y-intercepts and one x-intercept. It is not the graph of a function.

Here's one way to get there.

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Start with the parent function y = |x| and scale it down so that it has a y-intercept of -1 and x-intercepts at ±1.

Now, it is ...

  f(x) = |x| -1

We want to scale this vertically by a factor of -5. this puts the y-intercept at +5 and leaves the x-intercepts at ±1.

Horizontally, we want to scale the function by an expansion factor of 3. The transformed function g(x) will be ...

  g(x) = -5f(x/3) = -5(|x/3| -1) = -5/3|x| +5

This function has two x-intercepts at ±3 and one y-intercept at y=5. By swapping the x- and y-variables, we can get an equation for the graph you want:

  x = -(5/3)|y| +5

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Comment on this answer

Since there are no requirements on the graph other than it have the listed intercepts, you can draw it free-hand through the intercept points. It need not be describable by an equation.

step-by-step explanation:

i think something is missing in your question because y-1=(x-12) would not pass through the point (-4,-3). because if you replace x by -4 your are supposed to find y=-3 which is not the case her. if you replace x by -4 you find =-15.

i assume the right equation is y-1=1/4*(x-12) because this one is passing through both your points (if you are interested to know how i found that out i can tell you).

in this equation if i replace x by -4 i find y=-3 and if i replace x by 12, i find y=1 so the line representing this equation is passing through those 2 points.

then the standard form of the equation is finding an expression of y in function of x

y-1=1/4*(x-12)

y=x/4-3+1

y=x/4-2

hope that .

do you confirm there was a mistake in the question?

180-142=38 for the acute angle next to the entry

90+38=128

180-128=52

a=52

step-by-step explanation: