What is the perimeter of parallelogram ABCD?
1 answer:
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You could complete the square to state the vertex.
You could use the quadratic equation to find the roots (which are complex).
Try an example that will require both.
y = x^2 + 2x + 5
Step One
Get the graph. That's included below.
Step Two
Provide the steps for completing the square.
Note: we should get (-1,4)
y= (x^2 +2x ) + 5
y = (x^2 +2x + 1) + 5 - 1
y = (x +1)^2 + 4
The vertex is at (-1,4)
Step Three
Find the roots. Use the quadratic equation. Note that the graph shows us that the equation never crosses or touches the x axis. The roots are complex.
a = 1
b = 2
c = 5
x = -1 +/- 2i
x1 = -1 + 2i
x2 = -1 - 2i And we are done.
You could use the quadratic equation to find the roots (which are complex).
Try an example that will require both.
y = x^2 + 2x + 5
Step One
Get the graph. That's included below.
Step Two
Provide the steps for completing the square.
Note: we should get (-1,4)
y= (x^2 +2x ) + 5
y = (x^2 +2x + 1) + 5 - 1
y = (x +1)^2 + 4
The vertex is at (-1,4)
Step Three
Find the roots. Use the quadratic equation. Note that the graph shows us that the equation never crosses or touches the x axis. The roots are complex.
a = 1
b = 2
c = 5
x = -1 +/- 2i
x1 = -1 + 2i
x2 = -1 - 2i And we are done.