This equation is written in what's noted as slope/intercept type; y=mx+b, the place m is the slope of the line and b is the y-intercept. The y-intercept is the element at which x=0. subsequently, you have a element on your line already: (0, 2). The slope is the replace in y over the replace in x. To get yet another element, you upload the replace in y, 3, to the y of yet another element, and upload the replace in x, 4, to the x of the aforementioned "different element" on the line. because you recognize that (0, 2) is a element, upload 0 and four, and a pair of and 3. 0+4=4. 2+3=5. subsequently, (4, 5) is yet another element on the line. Then, merely draw a without postpone line connecting (0, 2) and (4, 5).
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Verified answer
shifting up or down is accomplished by adding an intercept to the right side.
shifting right or left is accomplished by adding or subtracting from the x before doing anything else.
To shift up 3 the equation becomes
y = X^2 +3
To shift up three and right 2 the equation becomes
y=(x-2)^2 +3
So when x is 2 in the new equation it acts like when x is zero in the old one.
This should give you the general idea.
This equation is written in what's noted as slope/intercept type; y=mx+b, the place m is the slope of the line and b is the y-intercept. The y-intercept is the element at which x=0. subsequently, you have a element on your line already: (0, 2). The slope is the replace in y over the replace in x. To get yet another element, you upload the replace in y, 3, to the y of yet another element, and upload the replace in x, 4, to the x of the aforementioned "different element" on the line. because you recognize that (0, 2) is a element, upload 0 and four, and a pair of and 3. 0+4=4. 2+3=5. subsequently, (4, 5) is yet another element on the line. Then, merely draw a without postpone line connecting (0, 2) and (4, 5).