Question 194437
Graph the parabola
1)y=(x-5)^2+4
Vertex form: (x-5)^2 = -(y-4)
Vertex at (5,4)
{{{graph(400,300,-10,10,-10,10,(x-5)^2 + 4)}}}
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2)y=4/3*^2 + 8x +12 
Question: Is the 1st term 4/9 ??
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3)Find the x-intercept and the coordinates of the vertex for the parabola 
y=x^2-6x-7.
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x-intercepts:
Let y = 0
x^2-6x-7 = 0
(x-7)(x+1) = 0
x = -1 or x = 7
These are the x-intercepts
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If there is more than one x intercept separate with commas.
x-intercept: (-1,0),(7,0)
vertex:
Complete the square on y = x^2-6x-7
x^2 - 6x + ? = y + 7 + ?
x^2 - 6x + 9 = y + 7 + 9
(x-3)^2 = y+16
Vertex: (3,-16)
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{{{graph(400,300,-10,10,-20,40,x^2-6x-7)}}}
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Cheers,
Stan H.