Identify the vertex, axis of symmetry, maximum or minimum and domain and range of each function 1. f(x)=4(x+2)^2-6 2. f(x)=10(x-1)^2+5

For first function vertex is (-2,-6), axis of symmetry is x = -2, minimum at (-2,-6), domain is all real numbers and range is . For second function vertex is (-1,5), axis of symmetry is x = -1, minimum at (-1,5), domain is all real numbers and range is .

Explanation:

The standard for of the parabola is,

Where (h,k) is vertex and vertex is the extreme point of a parabola. The axis of symmetry is y=h. If a>0 then f(x) is minimum at (h,k) and if a<0 then f(x) is maximum at (h,k).

The first function is,

On comparing this equation with standard form of parabola we get,

Therefore, the vertex is (-2,-6), axis of symmetry is x = -2, minimum at (-2,-6).

The function defined for all values of x, so the domain of f(x) is all real numbers. Since the minimum value of the function is -6, therefore range must be greater than or equal to -6.

The second function is,

On comparing this equation with standard form of parabola we get,

Therefore, the vertex is (-1,5), axis of symmetry is x = -1, minimum at (-1,5).

The function defined for all values of x, so the domain of f(x) is all real numbers. Since the minimum value of the function is 5, therefore range must be greater than or equal to 5.

option a, c,d are correct.

step-by-step explanation:

from the given figure, it is given that z is equidistant from the sides of the triangle rst, then from triangle tzb and triangle szb, we have

tz=sz(given)

bz=zb(common)

therefore, by rhs rule,δtzb ≅δszb

by cpctc, sz≅tz

also, from δctz and δasz,

tz=sz(given)

∠tcz=∠saz(90°)

by rhs rule, δctz ≅ δasz, therefore by cpctc, ∠ctz≅∠asz

also,from δasz and δzsb,

zs=sz(common)

∠zbs=∠saz=90°

by rhs rule, δasz ≅δzsb, therefore, by cpctc, ∠asz≅∠zsb

hence, option a, c,d are correct.

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step-by-step explanation: