Question 468755: Hi! I need help with two problems please? The only thing I am understanding is the minimum and maximum values. I cannot get the figures right.
Find the vertex, the line of symmetry, the maximum or minimum value of the quadratic function and graph the function.
f(x)=-2x^2+2x+4
f(x)= 1-x^2
Answer by lwsshak3(11628)   (Show Source):
You can put this solution on YOUR website! Find the vertex, the line of symmetry, the maximum or minimum value of the quadratic function and graph the function.
f(x)=-2x^2+2x+4
f(x)= 1-x^2
...
Standard form of parabola: y=ħA(x-h)^2+k, with (h,k) being the (x,y) coordinates of the vertex.
If the lead coefficient is>0, the parabola opens upwards and there is a minimum. If the lead coefficient is <0, the parabola opens downwards and there is a maximum. A is a multiplier which affects the steepness of the parabola.
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f(x)=-2x^2+2x+4
completing the square
f(x)=-2(x^2-x+1/4)+4+1/2
f(x)=-2(x-1/2)^2+9/2
This is a parabola which opens downwards with vertex at (1/2,9/2). (maximum=9/2)
Axis of symmetry: x=1/2
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f(x)= 1-x^2
f(x)=-x^2+1
Already in standard form,This is a parabola which opens downwards with vertex at (0,1). (maximum=1)
Axis of symmetry: x=0 or y-axis
See graph below as a visual check on answers
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