# How do you graph a quadratic function without plugging in points y= 2(x-3)^2+4?

Apr 15, 2015

The theory behind the graph transformation is simple.

If you have a graph of $y = F \left(x\right)$ then, to obtain a graph of $y = F \left(x + \epsilon\right)$, you have to shift the original graph to the left by $\epsilon$. If $\epsilon < 0$, it means, actually, shifting right by $| \epsilon |$.

Graph of $y = F \left(x\right) + \delta$ can be obtained from the original one by shifting it up by $\delta$. If $\delta < 0$ shift up by $\delta$ is, actually, a shift down by $| \delta |$.

Graph of $y = F \left(k \cdot x\right)$ (where $k$ is a constant) can be obtained from the original one by "squeezing" it horizontally towards $0$ by a factor of $| k |$ and, if $k < 0$, reflecting symmetrically relatively the Y-axis. Notice that for $| k | < 1$ "squeezing" is, actually, "stretching".

Graph of $y = k \cdot F \left(x\right)$ (where $k$ is a constant) can be obtained from the original one by "stretching" it vertically by a factor of $| k |$ and, if $k < 0$, reflecting symmetrically relatively the X-axis. Notice that for $| k | < 1$ "stretching" is, actually, "squeezing".

These graph manipulations are explained in all the details with proofs at Unizor in the item Algebra - Graphs - Manipulation.

Let's use these techniques for our problem.
To get to a graph of $y = 2 {\left(x - 3\right)}^{2} + 4$, we have to start from $y = {x}^{2}$, shift it to the right by $3$, stretch it vertically by a factor of $2$ and shift vertically by $4$.

The result is $y = 2 {\left(x - 3\right)}^{2} + 4$
graph{2(x-3)^2+4 [-5, 25, -5, 15]}
The bottom point of a parabola is at $\left(3 , 4\right)$ because we shifted it to the right by $3$ and up by $4$.
It intersects the Y-axis at
$y = 2 {\left(0 - 3\right)}^{2} + 4 = 22$