I found the answer to the question below, but I am not sure how it is supposed to differ from the one above?
Write an equation for a graph obtained by vertically stretching the
graph of y = x2 + 10 by a factor of 19.6, followed by a
vertical shift downward by 35 units.
Write an equation for a graph obtained by vertically shifting the graph of y = x^2 + 10 downward by 35 units,
followed by stretching the resulting graph by a factor of 19.6.
3 answers
(rewritten) I found the answer to the question below, but I am not sure how it is supposed to differ from the one above?
Write an equation for a graph obtained by vertically stretching the
graph of y = x^2 + 10 by a factor of 19.6, followed by a
vertical shift downward by 35 units.
Write an equation for a graph obtained by vertically stretching the
graph of y = x^2 + 10 by a factor of 19.6, followed by a
vertical shift downward by 35 units.
There's quite a difference between "shifting and stretching" and "stretching and shifting"
In the first case, the shift is also stretched.
For example, if you have a parabola y=x^2, if you stretch first, the graph still touches (0,0), no matter how far it is stretched. If it's then shifted, the stretched parabola is moved by that an=mount.
If it is shifted say, 5 units down, then after stretching by 3, the vertex is now 15 units down.
So, with that parabola,
stretch3-shift5: x^2 -> 3x^2 -> 3x^2-5
shift5-stretch3: x^2 -> x^2-5 -> 3(x^2-5) = 3x^2-15
Now apply that logic to your problem.
In the first case, the shift is also stretched.
For example, if you have a parabola y=x^2, if you stretch first, the graph still touches (0,0), no matter how far it is stretched. If it's then shifted, the stretched parabola is moved by that an=mount.
If it is shifted say, 5 units down, then after stretching by 3, the vertex is now 15 units down.
So, with that parabola,
stretch3-shift5: x^2 -> 3x^2 -> 3x^2-5
shift5-stretch3: x^2 -> x^2-5 -> 3(x^2-5) = 3x^2-15
Now apply that logic to your problem.
1 answer