All categories

3 years ago

What is a quadratic polynomial function with zeros 3 and -3?

Mathematics

1 answer:

Lemur [1.5K]3 years ago

8 0

Answer:

f(x) = x² - 9

Step-by-step explanation:

Given x = a is a zero of the polynomial f(x) then (x - a) is a factor

Given zeros are x = 3 and x = - 3, then corresponding factors are

(x - 3) and (x - (- 3) ), that is

(x - 3) and (x + 3)

The function is then the product of the factors, that is

f(x) = (x - 3)(x + 3) ← expand using FOIL

f(x) = x² - 9

You might be interested in

determine if you can use the HL congruence theorem to prove the triangles congruent. if not, tell what else you need to know

kherson [118]

I need an actual picture to solve this

4 0

3 years ago

IS THIS IN SIMPLIST FORM? PLEASE HELP! TY! <br><br> 2( x - 3 + 2 )

Phoenix [80]

Answer:

2x - 2

Step-by-step explanation:

There are two steps to take here. First add together the -3 + 2 in the brackets:

2(x - 3 + 2)

= 2(x - 1)

Now use the distributive property and multiply each term in the brackets by two:

2(x - 1)

= 2x - 2

And you now have it in simplest form.

3 0

3 years ago

A school bus can carry 66 pupils. how many pupils can ride on 12 such buses

denis-greek [22]

66x12=792 pupils can ride on 12 buses

4 0

3 years ago

HELP!! Algebra help!! Will give stars thank u so much <333

Anna35 [415]

Answers:

Part a) $\bf{\sqrt{x^2+(x^2-3)^2}$
Part b) 3
Part c) 2.24
Part d) 1.58

============================================================

Work Shown:

Part (a)

The origin is the point (0,0) which we'll make the first point, so let (x1,y1) = (0,0)

The other point is of the form (x,y) where y = x^2-3. So the point can be stated as (x2,y2) = (x,y). We'll replace y with x^2-3

We apply the distance formula to say...

$d = \sqrt{(x_1-x_2)^2+(y_1-y_2)^2}\\\\d = \sqrt{(0-x)^2+(0-y)^2}\\\\d = \sqrt{(0-x)^2+(-y)^2}\\\\d = \sqrt{x^2 + y^2}\\\\d = \sqrt{x^2 + (x^2-3)^2}\\\\$

We could expand things out and combine like terms, but that's just extra unneeded work in my opinion.

Saying $d = \sqrt{x^2 + (x^2-3)^2}$ is the same as writing d = sqrt(x^2-(x^2-3)^2)

-------------------------------------------

Part (b)

Plug in x = 0 and you should find the following

$d(x) = \sqrt{x^2 + (x^2-3)^2}\\\\d(0) = \sqrt{0^2 + (0^2-3)^2}\\\\d(0) = \sqrt{(-3)^2}\\\\d(0) = \sqrt{9}\\\\d(0) = 3\\\\$

This says that the point (x,y) = (0,3) is 3 units away from the origin (0,0).

-------------------------------------------

Part (c)

Repeat for x = 1

$d(x) = \sqrt{x^2 + (x^2-3)^2}\\\\d(1) = \sqrt{1^2 + (1^2-3)^2}\\\\d(1) = \sqrt{1 + (1-3)^2}\\\\d(1) = \sqrt{1 + (-2)^2}\\\\d(1) = \sqrt{1 + 4}\\\\d(1) = \sqrt{5}\\\\d(1) \approx 2.23606797749979\\\\d(1) \approx 2.24\\\\$

-------------------------------------------

Part (d)

Graph the d(x) function found back in part (a)

Use the minimum function on your graphing calculator to find the lowest point such that x > 0.

See the diagram below. I used GeoGebra to make the graph. Desmos probably has a similar feature (but I'm not entirely sure). If you have a TI83 or TI84, then your calculator has the minimum function feature.

The red point of this diagram is what we're after. That point is approximately (1.58, 1.66)

This means the smallest d can get is d = 1.66 and it happens when x = 1.58 approximately.

6 0

3 years ago

Look at the triangle: A right angle triangle is shown with hypotenuse equal to 10 centimeters. An acute angle of the triangle is

Ostrovityanka [42]

I hope this helps!!!!

4 0

3 years ago

Read 2 more answers

What is a quadratic polynomial function with zeros 3 and -3?​

What is a quadratic polynomial function with zeros 3 and -3?