Function = y=(x-4)^2-5+7calculate the x and y values for the vertex point of the parabola. state what the axis of symmetry is. set your quadratic function to zero (let y = 0) and use either the quadratic formula or completing the square method to solve for the two solutions of x when y = 0 calculate the discriminant for the quadratic equation.

Given equation of parabola is

Or we can simplify that to

This equaion looks similar to formula

Comparing both equation, we get:

a=1, h=4, k=2

(h,k) represents the vertex of the parabola

Hence vertex of the given parabola is (4,2).

So x-value of the vertex = 4

So y-value of the vertex = 2

Axis of symmetry is given by equation x=h

So the answer will be x=4

Now we will set y=0 and solve this using quadratic formula including desriminant

compare with quadratic equation  , we get:

a=1, b=-8, c=18

Descriminant is given by formula:

which is imaginary

Hence there will be no real solution for x-intercept.

answer: 20

step-by-step explanation: subtract the minimum date value from the maximum data value to find the date range. in this casem the date range is 42 - 22 = 20

hope this ! : ) ~zane

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.