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3 years ago

How many x or y intercepts can a quadratic function have?

Mathematics

1 answer:

gladu [14]3 years ago

4 0

Answer:

0-2 x-intercepts and 0-1 y-intercepts

Step-by-step explanation:

A quadratic function can have anywhere from 0 to 2 x-intercepts. This is possible because a function can have 0, 1, or 2 solutions.

A parabola can also have either 0 or 1 y intercepts. It is possible for the function to not intercept the y axis at all, and it is also possible for it to intercept it exactly one time.

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Minimize C = 6x + 3y Constraints: 4x + 3y > 24 4x + y = 16 6 x = 0, y > 0

vlabodo [156]

Answer:xy6

Step-by-step explanation:

4 0

3 years ago

Augusto has a rectangular garden that is 11 feet wide. The length of the diagonal is 1 foot greater than the length of the garde

brilliants [131]

The answer is C
call the length is a => the diagonal is a+1
11^2 + a^2=(a+1)^2
<=> 121 + a^2=a^2 + 2a +1
<=>120=2a
<=> a=60

5 0

3 years ago

Solve the equatuon x/2-(2/(x+1))=1 x=?

Snezhnost [94]

$\frac{x}{2}-\frac{2}{x+1}=1$

We have 2 denominators that we need to get rid of. Whenever there are the denominators, all we have to do is multiply all whole equation with the denominators.

Our denominators are both 2 and x+1. Therefore, we multiply the whole equation by 2(x+1)

$\frac{x}{2}[2(x+1)]-\frac{2}{x+1}[2(x+1)] = 1[2(x+1)]$

Then shorten the fractions.

$\frac{x}{2}[2(x+1)]-\frac{2}{x+1}[2(x+1)] = 1[2(x+1)]\\x(x+1)-2(2)=1(2x+2)$

Distribute in all.

$x^2+x-4=2x+2$

We should get like this. Because the polynomial is 2-degree, I'd suggest you to move all terms to one place. Therefore, moving 2x+2 to another side and subtract.

$x^2+x-4-2x-2=0\\x^2-x-6=0\\$

We are almost there. All we have to do is, solving for x by factoring. (Although there are more than just factoring but factoring this polynomial is faster.)

$(x-3)(x+2)=0\\x=3,-2$

Thus, the answer is x = 3, -2

7 0

3 years ago

If an = 2n - 3, then what is the fifth term of the sequence?

boyakko [2]

Answer:

Step-by-step explanation:

To find the fifth term in the sequence, plug in 5 for n.

a₅=2(5)-3

Then just solve

a₅=10-3

a₅=7

3 0

2 years ago

Read 2 more answers

How to solve 8/x^2-1=6/x+1

NARA [144]

You first need to cross multiply to get 8x+8=6x^2-6. Then make the function into 6x^2-8x-14=0 which can be simplified into 3x^2-4x-7=0 by dividing the equation by 2. Then you must split the middle(-4x) to get 3x^2+3x-7x-7=0. This simplifies to 3x(x+1)-7(x+1)=0 and then to (3x-7)(x+1)=0. From here you just solve for each root which you will find are x= -1 and 7/3

8 0

3 years ago