![\bf ~\hspace{10em}\textit{function transformations} \\\\\\ \begin{array}{llll} f(x)= A( Bx+ C)^2+ D \\\\ f(x)= A\sqrt{ Bx+ C}+ D \\\\ f(x)= A(\mathbb{R})^{ Bx+ C}+ D \end{array}\qquad \qquad \begin{array}{llll} f(x)=\cfrac{1}{A(Bx+C)}+D \\\\\\ f(x)= A sin\left( B x+ C \right)+ D \end{array} \\\\[-0.35em] ~\dotfill\\\\ \bullet \textit{ stretches or shrinks horizontally by } A\cdot B\\\\ \bullet \textit{ flips it upside-down if } A\textit{ is negative}\\ ~~~~~~\textit{reflection over the x-axis}](https://tex.z-dn.net/?f=%5Cbf%20~%5Chspace%7B10em%7D%5Ctextit%7Bfunction%20transformations%7D%20%5C%5C%5C%5C%5C%5C%20%5Cbegin%7Barray%7D%7Bllll%7D%20f%28x%29%3D%20A%28%20Bx%2B%20C%29%5E2%2B%20D%20%5C%5C%5C%5C%20f%28x%29%3D%20A%5Csqrt%7B%20Bx%2B%20C%7D%2B%20D%20%5C%5C%5C%5C%20f%28x%29%3D%20A%28%5Cmathbb%7BR%7D%29%5E%7B%20Bx%2B%20C%7D%2B%20D%20%5Cend%7Barray%7D%5Cqquad%20%5Cqquad%20%5Cbegin%7Barray%7D%7Bllll%7D%20f%28x%29%3D%5Ccfrac%7B1%7D%7BA%28Bx%2BC%29%7D%2BD%20%5C%5C%5C%5C%5C%5C%20f%28x%29%3D%20A%20sin%5Cleft%28%20B%20x%2B%20C%20%5Cright%29%2B%20D%20%5Cend%7Barray%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill%5C%5C%5C%5C%20%5Cbullet%20%5Ctextit%7B%20stretches%20or%20shrinks%20horizontally%20by%20%7D%20A%5Ccdot%20B%5C%5C%5C%5C%20%5Cbullet%20%5Ctextit%7B%20flips%20it%20upside-down%20if%20%7D%20A%5Ctextit%7B%20is%20negative%7D%5C%5C%20~~~~~~%5Ctextit%7Breflection%20over%20the%20x-axis%7D)
with that template in mind, let's see
down by 5 units, D = -5
to the left by 4 units, C = +4
The answer is 8.
8 x 3 = 24
24 + 1 = 25
Hope I helped :)
pumped up kicks
Step-by-step explanation:
pumped up kicks
Y = 3x + 3 this is equation 1
y = x -1 this is equation 2
since equation 1 and 2 already defined y in terms of x, we're just gonna substitute y into the other equation. as 2 is shorter I'm going with that today
x - 1 = 3x +3
-1 - 3 = 3x - x
-4 = 2x
x = -2
now we've got x just sub it in equation 2
y = -2 -1
y = -3
so the answer is (-2,-3)
Answer:
The formula for this quadratic function is x*2 +6x+13
Step-by-step explanation:
If we have the vertex and one point of a parabola it is possible to find the quadratic function by the use of this
y= a (x-h)*2 + K
Quadratic function looks like this
y= ax*2 + bx + c
So let's find the a
y= a (x-h)*2 + K where
y is 13, x is 0, h is -3 and K is 4
13= a (0-(-3))*2 +4
13=9a +4
9=9a
9/9=a
1=a
The quadratic function will be
y= 1(x+3)*2 + 4
Let's get the classic form
(x+3)*2 = (x+3)(x+3)
(x*2+3x+3x+9)
x*2 +6x+13
f(0) = 13
