Solving Quadratic Equations (Graphing) <Urgent!>
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Hello, Mark!
Plug in a variety of values for \(\displaystyle x\) and find the values of the function.
. . Let \(\displaystyle x\) run from -4 to +4.
\(\displaystyle x=\)-\(\displaystyle 4:\;(\)-\(\displaystyle 4)^2\,+\,(\)-\(\displaystyle 4)\,-\,6\;=\;6\)
\(\displaystyle x=\)-\(\displaystyle 3:\;(\)-\(\displaystyle 3)^2\,+\,(\)-\(\displaystyle 3)\,-\,6\;=\;0\)
\(\displaystyle x=\)-\(\displaystyle 2:\;(\)-\(\displaystyle 2)^2\,+\,(\)-\(\displaystyle 2)\,-\,6\;=\;\)-\(\displaystyle 4\)
\(\displaystyle x=\)-\(\displaystyle 1:\;(\)-\(\displaystyle 1)^2\,+\,(\)-\(\displaystyle 1)\,-\,6\;=\;\)-\(\displaystyle 6\)
\(\displaystyle x=\,0:\;0^2\,+\,0\,-\,6\;=\;\)-\(\displaystyle 6\)
\(\displaystyle x=\,1:\;1^2\,+\,1\,-\,6\;=\;\)-\(\displaystyle 4\)
\(\displaystyle x=\,2:\;2^2\,+\,2\,-\,6\;=\;0\)
\(\displaystyle x=\,3:\;3^2\,+\,3\,-\,6\;=\;6\)
\(\displaystyle x=\,4:\;4^2\,+\,4\,-\,6\;=\;14\)
This gives us a set of points: \(\displaystyle (\)-\(\displaystyle 4,6),\,(\)-\(\displaystyle 3,0),\,(\)-\(\displaystyle 2,\)-\(\displaystyle 4),\,(\)-\(\displaystyle 1,\)-\(\displaystyle 6),\,(0,\)-\(\displaystyle 6),\,(1,\)-\(\displaystyle 4),\,(2,0),\,(3,6),\,(4,14)\)
Plot the points.
The points where the graph crosses the x-axis are the solutions.
Therefore, the solutions are: \(\displaystyle \:x\,=\,-3\) and \(\displaystyle x\,=\,2\)
This is the most primitive method for solving an equation . . .
Plug in a variety of values for \(\displaystyle x\) and find the values of the function.
. . Let \(\displaystyle x\) run from -4 to +4.
\(\displaystyle x=\)-\(\displaystyle 4:\;(\)-\(\displaystyle 4)^2\,+\,(\)-\(\displaystyle 4)\,-\,6\;=\;6\)
\(\displaystyle x=\)-\(\displaystyle 3:\;(\)-\(\displaystyle 3)^2\,+\,(\)-\(\displaystyle 3)\,-\,6\;=\;0\)
\(\displaystyle x=\)-\(\displaystyle 2:\;(\)-\(\displaystyle 2)^2\,+\,(\)-\(\displaystyle 2)\,-\,6\;=\;\)-\(\displaystyle 4\)
\(\displaystyle x=\)-\(\displaystyle 1:\;(\)-\(\displaystyle 1)^2\,+\,(\)-\(\displaystyle 1)\,-\,6\;=\;\)-\(\displaystyle 6\)
\(\displaystyle x=\,0:\;0^2\,+\,0\,-\,6\;=\;\)-\(\displaystyle 6\)
\(\displaystyle x=\,1:\;1^2\,+\,1\,-\,6\;=\;\)-\(\displaystyle 4\)
\(\displaystyle x=\,2:\;2^2\,+\,2\,-\,6\;=\;0\)
\(\displaystyle x=\,3:\;3^2\,+\,3\,-\,6\;=\;6\)
\(\displaystyle x=\,4:\;4^2\,+\,4\,-\,6\;=\;14\)
This gives us a set of points: \(\displaystyle (\)-\(\displaystyle 4,6),\,(\)-\(\displaystyle 3,0),\,(\)-\(\displaystyle 2,\)-\(\displaystyle 4),\,(\)-\(\displaystyle 1,\)-\(\displaystyle 6),\,(0,\)-\(\displaystyle 6),\,(1,\)-\(\displaystyle 4),\,(2,0),\,(3,6),\,(4,14)\)
Plot the points.
Code:
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- - o - - - - - | - - - - o - -
(-3,0) | (2,0)
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|Therefore, the solutions are: \(\displaystyle \:x\,=\,-3\) and \(\displaystyle x\,=\,2\)
Thje vertex is at x = -b/2a (from ax²+bx+c) = -1/2 = -.5 so a dot at
(-.5,f(x)) = (-.5,-6.250)
It factors into (x+3)(x-2) so a dot at (-3,0) and (2,0)
That gives a good start.
f(-2) = -4 so a dot at (-2,-4)
f(1) = -4 so a dot at (1,-4)
f(-4) = 6 so a dot at (-4,6)
f(3) = 6 so a dot at (3,6)
That should be enough points to do the graph.
If you need more points do f(?) where x=? is where you think you want another point.
The solution was accidently found by factoring but it is where the graph crosses y=0, at x=-3 and 2
(-.5,f(x)) = (-.5,-6.250)
It factors into (x+3)(x-2) so a dot at (-3,0) and (2,0)
That gives a good start.
f(-2) = -4 so a dot at (-2,-4)
f(1) = -4 so a dot at (1,-4)
f(-4) = 6 so a dot at (-4,6)
f(3) = 6 so a dot at (3,6)
That should be enough points to do the graph.
If you need more points do f(?) where x=? is where you think you want another point.
The solution was accidently found by factoring but it is where the graph crosses y=0, at x=-3 and 2
.Follow the same steps as the first.
Use -b/2a
Factor -(x²+3x-4)=0
About half-way between the vertex and the zeros is usually good. In this case x=0 and x=-3 look good.
Then two on the other side of the x axis. x= -5 and 2 for instance.
Lizzie numbers are good but subtract the vertex x from them.
Use numbers that give a good idea of the shape.
Definitly the vertex. Zeros if it crosses the x axis. The x of any large gaps to keep the shape.
Use -b/2a
Factor -(x²+3x-4)=0
About half-way between the vertex and the zeros is usually good. In this case x=0 and x=-3 look good.
Then two on the other side of the x axis. x= -5 and 2 for instance.
Lizzie numbers are good but subtract the vertex x from them.
Use numbers that give a good idea of the shape.
Definitly the vertex. Zeros if it crosses the x axis. The x of any large gaps to keep the shape.