# Find the value of t and the column vector/matrix (p)(q)<---(p is the upper value and q is the lower value in the column vector)Let f(x) = 3(x + 1)^2 - 12 and g(x) = x^2. The graph of f may be...

Find the value of **t **and the column vector/matrix (p)(q)<---(p is the upper value and q is the lower value in the column vector)

Let f(x) = 3(x + 1)^2 - 12 and g(x) = x^2. The graph of **f** may be obtained from the graph of **g** by the two transformations: a strectch of scale factor **t** in the y-direction followed by a translation of a column/matrix (p)(q). <---( p is the upper value and q is the lower value in the column vector/matrix)

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### 1 Answer

You need to remember that f(x+c) means a shifting of the graph of f(x) to the left, by c units, hence,you may get the graph `f(x) = (x+1)^2` , starting from the graph of `g(x) = x^2` , shifting the graph of g(x) to the left, by 1 unit.

You need to remember that `3(x+1)^2` means a vertically stretch of the shifted graph `f(x)= (x+1)^2` by a factor of 3.

You need to remember that `f(x) = 3(x+1)^2 - 12` means a downward shift of the graph `f(x) = 3(x+1)^2` by 12 units.

**Hence, starting from the graph of `g(x) = x^2` , using two shifts, one to the left by p=1 unit and one downward by q=12 units and a vertically stretch by a factor of t=3, yields the graph of `f(x) = 3(x+1)^2 - 12` .**