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3 years ago

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which must be true of a quadratic function whose vertex is the same as its y-intercept? the axis of symmetry for the function is

x = 0. the axis of symmetry for the function is y = 0. the function has no x-intercepts. the function has 1 x-intercept.

2 answers:

Zina [86]3 years ago

8 0

Answer:

a. The axis of symmetry for the function is x = 0.

Step-by-step explanation:

edge 2020

Sholpan [36]3 years ago

5 0

If <u>quadratic function </u> $y=ax^2+bx+c$ has <u>vertex </u>that is the same as its <u>y-intercept</u>, then

$x_v=-\dfrac{b}{2a}=0,\\ \\b=0$

and its equation is $y=ax^2+c.$

The graph of this quadratic function is translated graph of the function $y=ax^2$ up or down (depends on c).

This means that there could be

two x-intercepts;
one x-intercept;
no x-intercepts.

The axis of symmetry of the graph of the function $y=ax^2$ is x=0 (y-axis).

Answer: correct choice is A

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Question 1,2, and 3 how do i factor those? Can you show the work and explain how?

DiKsa [7]

1: $3n^{2}+9n+6$

notice that each part is divisible by 3

$3n^{2}$ ÷ 3 = $n^{2}$

9n ÷ 3 = 3n

6 ÷ 3 = 2

so it becomes $3(n^{2} +3n+2)$

3n can be rewritten as 2n+n

-you want to rewrite it into two numbers that multiply to the number that's alone (in this case 2)

which would get you

$3(n^{2} +2n+n+2)$

Now that it's rewritten, you can factor out n + 2 from the equation.

<u><em>the answer is </em></u>

3(n+2)(n+1)

And you can check that by multiplying (n+2)(n+1) which is $n^{2} +2n+n+2$ and then each of those by 3, which is $3n^{2} +6n+3n+6$ or $3n^{2}+9n+6$ , our origional equation

2: $28+x^{2} -11x$

So I rewrote this as $x^{2} -11x+28$ (it's the same thing, just reordered using the commutative property)

now -11x can be rewritten as -4x-7x

(remember, the two numbers should multiply to equal 28, which is our constant.)

$x^{2} -4x-7x+28$

now we can factor out x from the first expression and -7 from the second

$x(x-4)-7(x-4)$

and lastly you factor out x-4,

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Make sure to check your work and make sure it multiplies to $x^{2} -11x+28$

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The first thing I notice when looking at this problem is that both 9 and 4 are perfect squares. Not only that, but they are the squares of 2 and 3, which are products of -12

So if you rewrite 9 as $3^{2}$ and 4 as $2^{2}$ , the equation becomes

$3^{2} x^{2} -12x+2^{2}$

now that $3^{2} x^{2}$ is ugly so it can be turned into $(3x)^{2}$

and -12x can be rewritten as $-2*3x*2$

so our equation now looks like $(3x)^2-2*3x*2+2^{2}$

There's a rule that says $a^{2} -2ab+b^{2} = (a-b)^{2}$

In our case, a=3x and b=2

<u><em>so the final answer is</em></u>

$(3x-2)^2$

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