Question

Graph the following quadratic functions, locating the vertices and x- intercepts (if any) accurately.

Answer 1

\[f (x) = (x-2)^2 +3\]

Answer 2

if the parabola is written as
f(x)=a(x-h)^2+k
the vertex is (h,k)
in your case, vertex is (2, 3)

Answer 3

to find the x-intercepts, expand (x-2)^2+3 to x^2-4x+7 and set it to 0
x^2-4x+7=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

Answer 4

Wait i got confused? How do i graph it?

Answer 5

do you know how to graph y=x^2?

Answer 6

I think so but i am not sure..

Answer 7

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

Answer 8

I wish you could draw graphs on here.

Answer 9

Can someone please show me how to graph this?

Answer 10

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)=

i hope you get an idea (i am sorry the drawing couldnt be better)