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Ymorist [56]
2 years ago
8

Write a quadratic equation for a parabola with roots at (-2, 0) & (4,0) and a y-intercept at ( 0, -16) Write your answer in

factored form
Mathematics
1 answer:
meriva2 years ago
8 0

9514 1404 393

Answer:

  y = 2(x +2)(x -4)

Step-by-step explanation:

The y-intercept will be a constant times the product of the roots. Here, the product of the roots is (-2)(4) = -8, so the constant of interest is -16/-8 = 2. That constant is the coefficient of the leading term of the quadratic, so is a multiplier of the factored form.

  y = 2(x +2)(x -4)

__

For root p, (x-p) is a factor in the factored form.

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7 0
3 years ago
Jane wishes to bake an apple pie for dessert. The baking instructions say that she should bake the pie in an oven at a constant
Viktor [21]

Answer:

Therefore k= \frac{ln2 }{18}, A=184

Step-by-step explanation:

Given function is

T(t)=230 -e^{-kt}

where T(t) is the temperature in °C and t is time in minute and A and k are constants.

She noticed that after 18 minutes the temperature of the pie is 138°C

Putting T(t) =138°C and t= 18 minutes

138=230 -Ae^{-k\times 18}

\Rightarrow  -Ae^{-18k}=138-230

\Rightarrow  Ae^{-18k}=92 .....(1)

Again after 36 minutes it is 184°C

Putting T(t) =184°C and t= 36 minutes

184=230-Ae^{-k\times 36}

\Rightarrow Ae^{-36k}=230-184

\Rightarrow Ae^{-36k}=46.......(2)

Dividing (2) by (1)

\frac{Ae^{-36k}}{Ae^{-18k}}=\frac{46}{92}

\Rightarrow e^{-18k}=\frac{46}{92}

Taking ln both sides

ln e^{-18k}=ln\frac{46}{92}

\Rightarrow -18k =ln (\frac12)

\Rightarrow -18k= ln1-ln2

\Rightarrow k= \frac{ln2 }{18}

Putting the value k in equation (1)

Ae^{-18\frac{ln2}{18}}=92

\Rightarrow A e^{ln2^{-1}}=92

\Rightarrow A.2^{-1}=92

\Rightarrow \frac{A}{2}=92

\Rightarrow A= 92 \times 2

⇒A= 184.

Therefore k= \frac{ln2 }{18}, A=184

7 0
3 years ago
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