1. if possible, find the equation for the axis of symmetry for the graph of a quadratic function with the given pair of
coordinates. if not possible, explain why.
a. (3, 10) (15, 10)
b. (−2, 6) (6,4)

Step-by-step explanation:

its almost always possible!

a)(3,10)(15,10)

note that y-coordinates of the points is the same. So we can visualize that both of these points lie on a single horizontal line.

A parabola that crosses both of these points must be along the y-axis as well.

The axis of symmetry lies in between these two points. So, what we need to do is only find the midpoint of the two points! let's denote the midpoint as

we don't really need to compute the y-coordinate of the midpoint since it is already known to be 10.

this point (9,10) lies on the axis of symmetry. since the parabola is along the y-axis. the axis of symmetry must be parallel to the y-axis. Hence the equation of the axis of symmetry is

sidenote:

funny thing is that axis of symmetry can also be y = 10, IF you think that the parabola 'squeezed' into a horizontal line. (but that's not gonna be a good answer, and it depends whether the answer is preferred or not, so i'm not gonna recommend this as an answer! Although, it's a good topic for discussion)

b. (-2, 6), (6,4)

we can find the midpoint of these points as well. Because the axis of symmetry always lies on the midpoint.

this time the axis of symmetry can either be along the x-axis or the y-axis since both are possible!

so, two answers are possible:

These are the possible equations of the axes of symmetry that can be formed by a parabola given the two coordinates.

sidenote:

The axis of symmetry always lies on the midpoints of the two coordinates! So its almost always possible that there will exist a parabola if only two coordinates are given (since only two coordinates are given, this allows us to be free in 'choosing' how the quadratic function would shape like in order cross any desired coordinate)

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.

go and ask your mom to you