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

Question

Avery Alexander · Accepted Answer

The graph transformation has a straightforward theory behind it. If you have a graph of #y=F(x)# then, to obtain a graph of #y=F(x+epsilon)#, 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(x)+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(k*x)# (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*F(x)# (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". The item Algebra - Graphs - Manipulation at Unizor provides a detailed explanation of these graph manipulation techniques along with proofs. Let's use these techniques for our problem.
To get to a graph of #y=2(x-3)^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(x-3)^2+4#
 graph{2(x-3)^2+4 [-5, 25, -5, 15]}
The bottom point of a parabola is at #(3,4)# because we shifted it to the right by #3# and up by #4#. 
It intersects the Y-axis at
#y=2(0-3)^2+4=22#

Jace Anderson · Answer

undefined To graph a quadratic function $ y = 2(x - 3)^2 + 4 $ without plugging in points, you can use the vertex form of a quadratic function, which is $ y = a(x - h)^2 + k $. In this form, the vertex of the parabola is at the point $ (h, k) $.

For the given function $ y = 2(x - 3)^2 + 4 $:
- The vertex is at $ (3, 4) $.
- Since the coefficient of $ x^2 $ is positive (2 in this case), the parabola opens upwards.
- You can use the vertex and the shape of the parabola to sketch the graph accurately.