What form of a quadratic function would be graphed having the vertex at the same point as the y-intercept

1 answer:

5 0

The functions would be in the form of ax2+c. The axis of symmetry would be the y-axis , or x=0 , because b would be zero. Likewise, the y-intercept is not important, as any value of c will still yield a vertex at the y-intercept.

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question is ,four children saved £29 to go on a coach. They each saved the same amount. How much did each saved

Andreas93 [3]

If they saved £29 all together, and if each of them saved the same amount,
then each saved £29/4 = £7.25 or £7 and 25p.

(Before decimalization in 1971, £7.25 would have been £7 and 5s .)

5 0

3 years ago

Read 2 more answers

Start time 3:30 pm end time 7:00 pm what is elapsed time?

DiKsa [7]

Hello there!

Your answer is 3 hours and 30 minutes.

I hope I have been of assistance!

~ Fire

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3 years ago

I need help please answer these two questions for me! Will give points.

jarptica [38.1K]

Answer:

7.) 7

10.) 0

Step-by-step explanation:

When it means "evaluate the function", it's in essence asking us to see what the function spits out when we feed it a certain input. Our inputs are our x values, which spit out a y value.

Evaluating the function when x = 1:

Let's look at where the function has an x value of 1. We see it near the bottom of the table and see the y value associated with the input is 7. So when the function is fed 1 as an input, it spits out 7.

Evaluating the function when f(x) = - 2:

This one is a weird because of the new notation. Just think of it as some value of f, which we don't know (so we represent it as an x-variable) must equal -2. So let's look at our table to find out where our output is -2. We find that when f(x) = -2 the input is 0. So the input which gives -2 is 0.

7 0

3 years ago

Find the integral using substitution or a formula.

Nadusha1986 [10]

$\rm \int \dfrac{x^2+7}{x^2+2x+5}~dx$

Derivative of the denominator:
$\rm (x^2+2x+5)'=2x+2$

Hmm our numerator is 2x+7. Ok this let's us know that a simple u-substitution is NOT going to work. But let's apply some clever Algebra to the numerator splitting it up into two separate fractions. Split the +7 into +2 and +5.

$\rm \int \dfrac{x^2+2+5}{x^2+2x+5}~dx$

and then split the fraction,

$\rm \int \dfrac{x^2+2}{x^2+2x+5}~dx+\int\dfrac{5}{x^2+2x+5}~dx$

Based on our previous test, we know that a simple substitution will work for the first integral: $\rm \quad u=x^2+2x+5\qquad\to\qquad du=2x+2~dx$

So the first integral changes,

$\rm \int \dfrac{1}{u}~du+\int\dfrac{5}{x^2+2x+5}~dx$

integrating to a log,

$\rm ln|x^2+2x+5|+\int\dfrac{5}{x^2+2x+5}~dx$

Other one is a little tricky. We'll need to complete the square on the denominator. After that it will look very similar to our arctangent integral so perhaps we can just match it up to the identity.

$\rm x^2+2x+5=(x^2+2x+1)+4=(x+1)^2+2^2$

So we have this going on,

$\rm ln|x^2+2x+5|+\int\dfrac{5}{(x+1)^2+2^2}~dx$

Let's factor the 5 out of the intergral,
and the 4 from the denominator,

$\rm ln|x^2+2x+5|+\frac54\int\dfrac{1}{\frac{(x+1)^2}{2^2}+1}~dx$

Bringing all that stuff together as a single square,

$\rm ln|x^2+2x+5|+\frac54\int\dfrac{1}{\left(\dfrac{x+1}{2}\right)^2+1}~dx$

Making the substitution: $\rm \quad u=\dfrac{x+1}{2}\qquad\to\qquad 2du=dx$

giving us,

$\rm ln|x^2+2x+5|+\frac54\int\dfrac{1}{\left(u\right)^2+1}~2du$

simplying a lil bit,

$\rm ln|x^2+2x+5|+\frac52\int\dfrac{1}{u^2+1}~du$

and hopefully from this point you recognize your arctangent integral,

$\rm ln|x^2+2x+5|+\frac52arctan(u)$

undo your substitution as a final step,
and include a constant of integration,

$\rm ln|x^2+2x+5|+\frac52arctan\left(\frac{x+1}{2}\right)+c$

Hope that helps!
Lemme know if any steps were too confusing.

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3 years ago

A scientist had 3/4 liter of solution. HE used 1/6 of the solution for an experiment..How much solution did the scientist use fo

Leviafan [203]

Answer:

$\frac{1}{8}\ liter$