Given the quadratic function f (x), the vertex of its image is (- 1,2) and passes through the origin to find f (x)

Given the quadratic function f (x), the vertex of its image is (- 1,2) and passes through the origin to find f (x)

Vertex is (- 1,2)
f(x)=a(x+1) ²+ two
If it crosses the origin, f (0) = 0
So a (0 + 1) ²+ 2=0
a=-2
f(x)=-2x ²- 4x

Given the quadratic function f (x), its image vertex is (1,2) and passes through the coordinate origin, then f (x)=

The image vertex of quadratic function f (x) is (1,2), so f (x) = a (x-1) ^ 2 + 2
And it passes through the coordinate origin, that is, f (0) = a (0-1) ^ 2 + 2 = 0
Get a = - 2
That is, f (x) = - 2 (x-1) ^ 2 + 2 = - 2x ^ 2 + 4x

The left and right limits of function f (x) = |x| / X at X. = 0 are studied, and whether there is a limit is pointed out

When x tends to 0 from left, f (x) = - 1, so limf (x) = - 1; When x tends to 0 from right, f (x) = 1, so the left and right limits at limf (x) = 1.0 are not equal, so the limit at 0 does not exist

Function limit: X of sin2 (x-1) / X-1 tends to the limit of 1, and 2 after sin is square

t=x-1
sin2t/t=2sintcost/t
t==>0,c0st==>1
sint/t ==1
The limit is 2

When x approaches zero, find the limit of one square part of sin minus one square part of X

When (x → 0), the original limit = 1 / (SiNx) ^ 2-1 / x ^ 2 = [x ^ 2 - (SiNx) ^ 2] / x ^ 4 = (2x-sin2x) / (4x ^ 3) = (2-2cos2x) / 12x ^ 2 (two applications of lobitar's law) = 2 * (2x ^ 2) / (12x ^ 2) {(x → 0), 1-cos2x is equivalent to (2x) ^ 2 / 2 = 2x ^ 2} = 1 / 3

Find the function limit LIM (x →∞) xsinxsin1 / x ^ 2 (1 / 2 of the square of x)

lim(x→∞) xsinxsin1/x^2
=lim(x→∞) (1/x)sinx[sin1/x^2]/[1/x^2]
=lim(x→∞) (1/x)sinx
=0