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Suppose a parabola has a vertex (-8,-7) and also passes through the point (-7,-4). write the equation of the parabola in vertex form.

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The answer is y=3(x+8) -7 .. I just dont know how to get it
Vertex form is: \[a(x-h)^{2}+k\]

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Other answers:

\[(h,k) \] is your vertex
I know that but when I plug the problem into it I don't get the correct answer. What am I doing wrong. Can you please explain how to do it to me
So -8 and -7 are vertex
-8 = h -7= k
a ( -7+8) Squared + 7 is that right
Subtract -7 from both sides of my equation.
Okay. So now do we have to plug in the other points that it passes through
Then do I add 4 to the 7 ?
No, because you want to solve for a. SO add the 7 to the 4
which is 3
ahhh I get it now! Thank you so much for your help. Changing topics can you help me with another problem?
What point in the feasible region minimizes the objective function? X > 0 y>0 y<4 Y < -x +5 Above are the constraints Objective Function: C= -9x+y The answer is (5,0) but I don't understand how to get it.
gimme a sec please
okay no problem!
I was thinking you had to plug some stuff in to get the final answer
I used my graphing calc. I graphed y=-x+5
the x intercept of that is (5,0)
That's the vertex.. right
Cause have another example of a problem like that.. & that doesn't work Graph the system of constraints and find the value of x and y that minimize the objective function. 20. Constraints x ≥ 0 y ≥ 0 y ≤ 3 y < − 2x + 5 Ï Ì Ó Objective function: C = −8x + 5y
(5,0) is the vertex because y=-x+5 is a line
please ignore those random letters the answer is (5 / 2,0)
ohh i see what your saying
If you graph, y=-2x+5, your x intercept is (5/2, 0)
how do you see that..
I put it in my graphing calculator

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