Take a look at teen 1's initial polynomials for three layers of hills. teen 1 has described what he thinks each modification to the initial polynomial will do. sketch each curve (in a different color, preferably). (you will be asked to find and repair mistakes in the next section.) (6 points: 2 points for each curve) hill 1: my first layer needs three peaks. i chose three points on the x-axis (the three zeros), and then i wrote each as a binomial to create the first polynomial. f(x) = (x – 1)(x – 3)(x – 4) hill 2: my second layer should be a useful transformation of the first. using the first polynomial as the base, i multiplied it by to flip it and make the hills steeper. then, i added 3 to shift it to the right. f(x) =– (x – 1)(x – 3)(x – 4) + 3 hill 3: this hill should be a shallow parabola that will rise up in the foreground of my picture. f(x) = 4(x–)(x – 5)

This question is based on the graph. Therefore, the errors are:

(1) Error is that, three zeros (x-intercepts) does not make 3 peaks (vertices).

(2) Error is that, multiplying it by a negative is a reflection (not a steeper slope). Adding three on the outside of the factors is a vertical shift up (not a horizontal shift to the right).

(3) Error is that, multiplying by four makes a steeper slope (not a shallow slope).

Given:

Take a look at teen 1's initial polynomials for three layers of hills. Teen 1 has described what he thinks each modification to the initial polynomial will do.

According to the question,

Attachment of each curve, is below.

(1) I chose three points on the x-axis (the three zeros), and then I wrote each as a binomial to create the first polynomial. F(x) = (x – 1)(x – 3)(x – 4).

Error is that, three zeros (x-intercepts) does not make 3 peaks (vertices).

(2) Using the first polynomial as the base, I multiplied it by to flip it and make the hills steeper. Then, I added 3 to shift it to the right. F(x) =– (x – 1)(x – 3)(x – 4) + 3.

Error is that, multiplying it by a negative is a reflection (not a steeper slope). Adding three on the outside of the factors is a vertical shift up (not a horizontal shift to the right).

(3)This hill should be a shallow parabola that will rise up in the foreground of my picture. F(x) = 4(x–)(x – 5).

Error is that, multiplying by four makes a steeper slope (not a shallow slope).

Therefore, the errors are in each question,

(1) Error is that, three zeros (x-intercepts) does not make 3 peaks (vertices).

(2) Error is that, multiplying it by a negative is a reflection (not a steeper slope). Adding three on the outside of the factors is a vertical shift up (not a horizontal shift to the right).

(3) Error is that, multiplying by four makes a steeper slope (not a shallow slope).

For more details, prefer this link:

link

See attachment for graphs

1) Error: 3 zeros (x-intercepts) does not make 3 peaks (vertices)

2) Errors: Multiplying it by a negative is a reflection (not a steeper slope). Adding 3 on the outside of the factors is a vertical shift up (not a horizontal shift to the right).

3) Error: Multiplying by 4 makes a steeper slope (not a shallow slope).

what are the statements

step-by-step explanation:

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.