Need answer asap algebra 2 math

Mathematics

1 answer:

Scorpion4ik [409]3 years ago

4 0

Answer:

Step-by-step explanation:

f(x) = 2x^2 - 4x + 5 Put brackets around the first 2 terms.

f(x) = (2x^2 - 4x) + 5 Take out the common factor of 2

f(x) = 2(x^2 - 2x) + 5 1/2 the second term and square.

f(x) = 2(x^2 -2x + (2/2)^2) + 5 Express brackets as a perfect square.

f(x) = 2(x - 1)^2 + 5 Multiply 2 * 1 (the term without an x. Subtract.

f(x) = 2(x - 1)^2 + 5 - 2 Combine

f(x) = 2(x - 1)^2 + 3

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Help with 30 please. thanks.

Svet_ta [14]

Answer:

See Below.

Step-by-step explanation:

We have the equation:

$\displaystyle y = \left(3e^{2x}-4x+1\right)^{{}^1\! / \! {}_2}$

And we want to show that:

$\displaystyle y \frac{d^2y }{dx^2} + \left(\frac{dy}{dx}\right) ^2 = 6e^{2x}$

Instead of differentiating directly, we can first square both sides:

$\displaystyle y^2 = 3e^{2x} -4x + 1$

We can find the first derivative through implicit differentiation:

$\displaystyle 2y \frac{dy}{dx} = 6e^{2x} -4$

Hence:

$\displaystyle \frac{dy}{dx} = \frac{3e^{2x} -2}{y}$

And we can find the second derivative by using the quotient rule:

$\displaystyle \begin{aligned}\frac{d^2y}{dx^2} & = \frac{(3e^{2x}-2)'(y)-(3e^{2x}-2)(y)'}{(y)^2}\\ \\ &= \frac{6ye^{2x}-\left(3e^{2x}-2\right)\left(\dfrac{dy}{dx}\right)}{y^2} \\ \\ &=\frac{6ye^{2x} -\left(3e^{2x} -2\right)\left(\dfrac{3e^{2x}-2}{y}\right)}{y^2}\\ \\ &=\frac{6y^2e^{2x}-\left(3e^{2x}-2\right)^2}{y^3}\end{aligned}$