The solution of a system of linear equations is where the two lines intersect....
2 answers:
Answer:
Step-by-step explanation:
True.
There is only 1 intersection point. That is the point where the x and y values are the same for both equations. For example if you have
y = 2x + 5 and
y = 3x + 2
The intersect point is (3,11). If you put 3 in each equation for x, they will return 11 for the y value.
I've included a graph to show you that this is true.
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Since you haven't provided the graph, I'll explain each one and you choose the one suiting your given.
<u>The parent modulus function is:</u>
g(x) = |x|
It is centered at the origin and opens upwards.
A coefficient inside the modulus |x+k| means that the function is shifted along the x-axis
If "k" is <u>positive</u>, the shift will be to the <u>left</u>. If "k" is <u>negative</u>, the shift will be to the <u>right</u>.
A coefficient outside the modulus |x| + h means that the function is shifted along the y-axis
If "h" is <u>positive</u>, the shift will be <u>upwards</u>. If "h" is <u>negative,</u> the shift will be <u>downwards</u>.
<u>Now, let's check each of the options:</u>
<u>g(x) = |x+4| - 2 :</u>
This function is shifted 4 units to the left and 2 units down. It will be centered at (-4,-2). Check the blue graph in the attachment.
<u>g(x) = |x-4| - 2 :</u>
This function is shifted 4 units to the right and 2 units down. It will be centered at (4,-2). Check the black graph in the attachment.
<u>g(x) = |x-2| - 4 :</u>
This function is shifted 2 units to the right and 4 units down. It will be centered at (2,-4). Check the red graph in the attachment.
<u>g(x) = |x-2| + 4 :</u>
This function is shifted 2 units to the right and 4 units up. It will be centered at (2,4). Check the green graph in the attachment.
All 4 graphs are shown in the attached picture.
Hope this helps :)
