The domain of y=ax^2 is all real numbers. Describe the range when (a) a>0 and (b) a<0.
1 answer:
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Write the vertex form of the equation and find the necessary coefficient to make it work.
.. y = a*(x +3)^2 -2
.. = ax^2 +6ax +9a -2
You require the y-intercept to be 7. So, for x=0, you have
.. 9a -2 = 7
.. 9a = 9
.. a = 1
The equation you seek is
.. y = x^2 +6x +7
.. y = a*(x +3)^2 -2
.. = ax^2 +6ax +9a -2
You require the y-intercept to be 7. So, for x=0, you have
.. 9a -2 = 7
.. 9a = 9
.. a = 1
The equation you seek is
.. y = x^2 +6x +7
Answer:
2
Step-by-step explanation:
because
So,