Graph the following quadratic functions, locating the vertices and x- intercepts (if any) accurately.
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\[f (x) = (x-2)^2 +3\]
if the parabola is written as
the vertex is (h,k)
in your case, vertex is (2, 3)
to find the x-intercepts, expand (x-2)^2+3 to x^2-4x+7 and set it to 0
solve for x
since the discriminant, b^2-4ac<0 in this case, there are no x-intercepts because the solution to this quadratic equation has 2 complex roots
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Wait i got confused? How do i graph it?
do you know how to graph y=x^2?
I think so but i am not sure..
if you can graph y=x^2, then you can graph y=(x-2)^2+3
y=(x-2)^2+3 is the graph of y=x^2 shifted right 2 units and up 3 units
I wish you could draw graphs on here.
Can someone please show me how to graph this?
yes you can. first find the vertex of the curve. its (-b/2a , -D/4a)
so that makes it (2,-3). next find the roots. they are none ofcourse. so, curve a symmetric parabola with vertex (2,3). and find out a few cordinates (like y-coordinates) for accuracyof the plot (ofcourse it will be free hand)=|dw:1333134848249:dw|
i hope you get an idea (i am sorry the drawing couldnt be better)