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3 years ago

11

Solve the following quadratic equation by factoring:

y^{2}" alt="y^{2}" align="absmiddle" class="latex-formula">-8y=0

1 answer:

Furkat [3]3 years ago

4 0

Answer:

y = 8 or y = 0

Step-by-step explanation:

Solve for y over the real numbers:

y^2 - 8 y = 0

Factor y from the left hand side:

y (y - 8) = 0

Split into two equations:

y - 8 = 0 or y = 0

Add 8 to both sides:

Answer: y = 8 or y = 0

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Anna [14]

Answer:

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Step-by-step explanation:

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Savatey [412]

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3 years ago

Find the absolute maximum and absolute minimum values, if any, of the function. (If an answer does not exist, enter DNE.)f(x) =

lyudmila [28]

Answer:

The absolute minimum value is " $-\frac{21}{4}$ " and the absolute maximum value is " $15$ ".

Step-by-step explanation:

Given:

$f(x)=x^2-x-5$

on,

$[0,5]$

By differentiating it, we get

⇒ $f'(x)=2x-1$

Set $f'(x)=0$

then,

⇒ $2x-1=0$

$2x=1$

$x=\frac{1}{2}$ (Critical point)

When x=0,

⇒ $f(x)=-5$

When $x=\frac{1}{2}$ ,

⇒ $f(x)=-\frac{21}{4}$ (Absolute minimum)

When $x=5$

⇒ $f(x)=15$ (Absolute maximum)

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2 years ago