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3 years ago

Find a polynomial function whose graph passes through (6,13),(10,-10) and (0,5)

Mathematics

2 answers:

dedylja [7]3 years ago

8 0

Answer:

See below

Step-by-step explanation:

Let the function be quadratic as the given points are not collinear, not a linear function

Standard form of quadratic equation:

y = ax^2 + bx + c

Substituting the coordinates, we get:

13 = a(6)^2 + b(6) + c
-10 = a(10)^2 + b(10) + c
5 = a(0)^2 + b(0) + c

From equation 3 we get c = 5, considering this in the other equations and simplifying:

13 = 36a + 6b + 5 ⇒ 36a + 6b = 8 ⇒ 18a + 3b = 4
-10 = 100a + 10b + 5 ⇒ 100a + 10b = -15 ⇒ 20a + 2b = -3

Subtract 3 times the second equation from 2 times the first:

2(18a) + 2(3b) - 3(20a) - 3(2b) = 4(2) - 3(-3)
36a - 60a = 8 + 9
-24a = 17
a = - 17/24

Finding the value of b:

18(-17/24) + 3b = 4
3b = 4 + 51/4
b = 67/12

The quadratic function is:

y = -17/24x^2 + 67/12x + 5

So the graph of any polynomial function:

y = -17/24x^2 + 67/12x + 5 + g(x)(x-6)(x-10)(x-0)

will pass through the given 3 points

svetlana [45]3 years ago

5 0

Answer:

Find a polynomial function whose graph passes through (6,13), (9,-11), (0,5)

1 Answers

Assuming a quadratic, we have that

y = ax^2 + bx + c

Since (0,5) is on the graph, c =5

And we have the remaining system

a(9)^2 + b(9) + 5 = -11

a(6)^2 + b(6) + 5 = 13 simplify

81a + 9b = -16 multiply through by 6 ⇒ 486a + 54b = - 96 (1)

36a + 6b = 8 multiply through by -9 ⇒ -324a -54b = -72 (2)

Add (1) and (2)

162a = -168

a = -28/27

To find b we have

36 (-28/27) + 6b = 8

-112/3 + 6b = 8

⇒ b = 68/9

The function is

y = - (28/27)x^2 + (68/9)x + 5

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3 years ago

The amount of syrup that people put on their pancakes is normally distributed with mean 63 mL and standard deviation 13 mL. Supp

andreyandreev [35.5K]

Answer:

(a) X ~ N( $\mu=63, \sigma^{2} = 13^{2}$ ).

$\bar X$ ~ N( $\mu=63,s^{2} = (\frac{13}{\sqrt{43} } )^{2}$ ).

(b) If a single randomly selected individual is observed, the probability that this person consumes is between 61.4 mL and 62.8 mL is 0.0398.

(c) For the group of 43 pancake eaters, the probability that the average amount of syrup is between 61.4 mL and 62.8 mL is 0.2512.

(d) Yes, for part (d), the assumption that the distribution is normally distributed necessary.

Step-by-step explanation:

We are given that the amount of syrup that people put on their pancakes is normally distributed with mean 63 mL and a standard deviation of 13 mL.

Suppose that 43 randomly selected people are observed pouring syrup on their pancakes.

(a) Let X = amount of syrup that people put on their pancakes

The z-score probability distribution for the normal distribution is given by;

Z = $\frac{X-\mu}{\sigma}$ ~ N(0,1)

where, $\mu$ = mean amount of syrup = 63 mL

$\sigma$ = standard deviation = 13 mL

So, the distribution of X ~ N( $\mu=63, \sigma^{2} = 13^{2}$ ).

Let $\bar X$ = sample mean amount of syrup that people put on their pancakes

The z-score probability distribution for the sample mean is given by;

Z = $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ ~ N(0,1)

where, $\mu$ = mean amount of syrup = 63 mL

$\sigma$ = standard deviation = 13 mL

n = sample of people = 43

So, the distribution of $\bar X$ ~ N( $\mu=63,s^{2} = (\frac{13}{\sqrt{43} } )^{2}$ ).

(b) If a single randomly selected individual is observed, the probability that this person consumes is between 61.4 mL and 62.8 mL is given by = P(61.4 mL < X < 62.8 mL)

P(61.4 mL < X < 62.8 mL) = P(X < 62.8 mL) - P(X $\leq$ 61.4 mL)

P(X < 62.8 mL) = P( $\frac{X-\mu}{\sigma}$ < $\frac{62.8-63}{13}$ ) = P(Z < -0.02) = 1 - P(Z $\leq$ 0.02)

= 1 - 0.50798 = 0.49202

P(X $\leq$ 61.4 mL) = P( $\frac{X-\mu}{\sigma}$ $\leq$ $\frac{61.4-63}{13}$ ) = P(Z $\leq$ -0.12) = 1 - P(Z < 0.12)

= 1 - 0.54776 = 0.45224

Therefore, P(61.4 mL < X < 62.8 mL) = 0.49202 - 0.45224 = 0.0398.

(c) For the group of 43 pancake eaters, the probability that the average amount of syrup is between 61.4 mL and 62.8 mL is given by = P(61.4 mL < $\bar X$ < 62.8 mL)

P(61.4 mL < $\bar X$ < 62.8 mL) = P( $\bar X$ < 62.8 mL) - P( $\bar X$ $\leq$ 61.4 mL)

P( $\bar X$ < 62.8 mL) = P( $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ < $\frac{62.8-63}{\frac{13}{\sqrt{43} } }$ ) = P(Z < -0.10) = 1 - P(Z $\leq$ 0.10)

= 1 - 0.53983 = 0.46017

P( $\bar X$ $\leq$ 61.4 mL) = P( $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ $\leq$ $\frac{61.4-63}{\frac{13}{\sqrt{43} } }$ ) = P(Z $\leq$ -0.81) = 1 - P(Z < 0.81)

= 1 - 0.79103 = 0.20897

Therefore, P(61.4 mL < X < 62.8 mL) = 0.46017 - 0.20897 = 0.2512.

(d) Yes, for part (d), the assumption that the distribution is normally distributed necessary.

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3 years ago

I need this problem solved real quick

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Let's take this problem step-by-step:

The ;unlabeled angle' adjacent to the 'outside angle of measure 78°'

⇒ is on a straight line

⇒ sum of angle measure = 180 degrees

$unlabeled_.angle+78 = 180\\unlabeled_.angle = 102$

Now we know:

⇒ sum of all angles in a triangle ⇒ 180°

Let's put that in equation form and solve:

$x + 102 + 57=180\\x + 159=180\\x = 21$

Answer: 21°

Hope that helps!

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Find a polynomial function whose graph passes through ​(6,13),(10,-10) and (0,5)

Find a polynomial function whose graph passes through (6,13),(10,-10) and (0,5)