Question 769251: f(x)=1/2(x+1)^2-3
how i find the points (x,y) in order to graph this parabola?
Found 2 solutions by josgarithmetic, lwsshak3:
Answer by josgarithmetic(39621)   (Show Source):
You can put this solution on YOUR website! You mean this: 
The function is in standard form allowing you to read the vertex and understand from lead coefficient greater than zero, vertex is a minimum. The vertex is at (-1,-3), and parabola opens upward. Setting x=0, y intercept is  , or the point, (0,-5/2).
Also, being in standard form makes find x intercepts fairly easy, because setting f(x)=0, just so few steps needed to solve for x.
 , and sqrt(6) is near 2.45.
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Note that the reason for  showing the zero is so that the rendering would work and the expression, equation be more easily read.
Answer by lwsshak3(11628) (Show Source):
You can put this solution on YOUR website! f(x)=
how i find the points (x,y) in order to graph this parabola?
This is an equation of a parabola that opens upwards.(lead coefficient>0)
Its standard form: y=A(x-h)^2+k, (h,k)=(x,y) coordinates of the vertex. A is a coefficient that affects the slope or narrowness of the curve.
With given equation, you have coordinates of the vertex (-1,-3)
Next, find the x-intercepts.
set y=0
1/2(x+1)^2-3=0
1/2(x+1)^2=3
(x+1)^2=6
(x+1)=±√6
x=-1±√6=-1±2.45
x=-3.45 and 1.45
You can now draw graph of given parabola with coordinates of the vertex, the x-intercepts, and knowing that the parabola opens upward.
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