SOLUTION: For each quadratic function (a) find the vertex and the axis of symmetry and (b) graph the function.
g(x)=x^2+5x+4.
a=1, b=5 and c=4
I started doing the problem and here is wh
Algebra ->
Quadratic-relations-and-conic-sections
-> SOLUTION: For each quadratic function (a) find the vertex and the axis of symmetry and (b) graph the function.
g(x)=x^2+5x+4.
a=1, b=5 and c=4
I started doing the problem and here is wh
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Question 280228: For each quadratic function (a) find the vertex and the axis of symmetry and (b) graph the function.
g(x)=x^2+5x+4.
a=1, b=5 and c=4
I started doing the problem and here is what I have so far.
x = -b/2*a
x= -5/2*1 = -5/2 or -2.5
g(1)= 1*(1)^2 + 5(1) + 4
g(1)=1 + 5 +4
So the vertex is (-2.5, 10).
Now I am confuse on how to get the axis of symmetry.
g(1)= 10 Answer by stanbon(75887) (Show Source):
You can put this solution on YOUR website! For each quadratic function (a) find the vertex and the axis of symmetry and (b) graph the function.
g(x)=x^2+5x+4.
a=1, b=5 and c=4
I started doing the problem and here is what I have so far.
x = -b/2*a
x= -5/2*1 = -5/2 or -2.5
That is the x-coordinate of the vertex.
Now find the y-coordinate:
g(-2.5)= 1*(-2.5)^2 + 5(-2.5) + 4
g(-2.5)= -2.25
So the vertex is (-2.5,-2.25).
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The axis of symmetry passes thru the vertex.
The parabola opens up so the axis is a vertical line.
So, the equation of the axis must be x=-2.5
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Let's see:
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Cheers,
Stan H.
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