Explorations of Second Degree Equations
By: Laura Lowe
Problem:
Graph the
parabola y = 2x^{2} + 3x  4
i.
Overlay a
new graph replacing each x by (x 
4).
ii. Change the equation to move the vertex of
the graph into the second quadrant.
iii. Change the equation to produce a graph
concave down that shares the same vertex.
Before graphing y = 2x^{2}
+ 3x – 4 we can deduce some
characteristics from the equation.
Since a > 0, we know
that the graph is concave up. We
also know the yintercept is (0,
4) because when a parabola is in standard form, as this on is, the yintercept is (0, c). If we
combine these two pieces of information, we can deduce that the function will
have 2 xintercepts. We can also fine the vertex using
x = b/2a and we get (0.75, 5.125). Now graph the parabola.



First we are asked to overlay
the original graph with a new graph that has each x replaced with (x – 4). We might
expect that this would move the graph 4 units to the left.


Replacing each x with (x
– 4) has moved the graph 4 units to the right. Why is
this? Look at this numerical
example.
Let x = 1.
Original function: y = 2(1)^{2} +3(1) – 4 = 1
New function: y = 2(1 –
4)^{2} + 3(4 – 1) – 4 = 2(–3)^{2}
+3(–3) – 4 = 5.
So when we subtract 4 from 1,
we end up with the same yvalue as
if we had plugged in –3 in the original function. LetŐs look at this on the graph.
So subtracting 4 takes the yvalues from four units from the left, thereby moving
the function to the right.
Now, the question asks us to
move the vertex of the function to the second quadrant. Since the yvalue of the vertex is 5.125, we will have to move
the vertex, and with it the entire function, up more than 5.125 units. LetŐs move the graph 8 units. My first thought is to try adding 8 to y. This is the same as subtracting 8 from
the right side of a function in standard form.


Again, exactly the opposite
happened. So, what if we subtract 8 from y? (Add 8 to the right.)


That worked. Why? Try plugging in numbers like we did before and see if you
can figure it out.
Last we are asked to graph
concave down that shares the same vertex.
We know for a graph to be concave down a < 0.
So we can change y = 2x^{2} + 3x – 4 to y = –2x^{2} + 3x – 4.


That has changed the function
to concave down, but it has also moved our vertex. The new vertex is (0.75, 2.875). So we need to move the function left 1.5 units and down 2.25
units. From what we learned
earlier, that means we should add
1.5 to each x and add 2.25 to y. When we expand and simplify
we get y = –2x^{2} – 3x – 6.25.
Now graph.


That is exactly what we
wanted.