Evaluate f(x) = –x2 + 1 for x = –3.

Mathematics

1 answer:

Katarina [22]2 years ago

3 0

The answer to this

is -8

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${2}^{x} - {3}^{x} = \sqrt{ {6}^{x} } - \sqrt{ {9}^{x } } \\ \\ {2}^{x} - {3}^{x} = \sqrt{ {6}^{x} } - \sqrt{ {( {3}^{2}) }^{x } } \\ \\ {2}^{x} - {3}^{x} = \sqrt{ {6}^{x} } - \sqrt{ {({3)}^{2x}}}\\ \\ {2}^{x} - \cancel{{3}^{x}} = \sqrt{ {6}^{x} } - \cancel{ {3}^{x} } \\ \\ {2}^{x} = \sqrt{ {6}^{x} } \\ squaring \: both \: sides \\ \\ {( {2}^{x} )}^{2} = {(\sqrt{ {6}^{x} })}^{2} \\ \\ {2}^{2x} = {6}^{x} \\ \\ {2}^{2x} = {(2 \times 3)}^{x} \\ \\ {2}^{2x} = {2 ^{x} \times 3}^{x} \\ \\ \frac{ {2}^{2x} }{ {2}^{x} } = 3^{x} \\ \\ {2}^{2x - x} = {3}^{x} \\ \\ {2}^{x} = {3}^{x} \\ \\ \frac{ {2}^{x} }{ {3}^{x} } = 1 \\ \\ \bigg( \frac{2}{3} \bigg)^{x} = 1 \\ \\ \implies \: x = 0 \\ \\ \because \: for \: x = 0 \\ \\ \bigg( \frac{2}{3} \bigg)^{0} = 1$

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2 years ago

What is the vertex of the graph y = x^2 + 6x - 5 ?

Dafna11 [192]

Given: <span>y = x^2 + 6x - 5. Then a = 1, b = 6 and c = -5.

The x-coord. of the vertex is given by x = -b / (2a), which here is x = -6 / (2*1) = -3.

Use the given formula </span><span>y = x^2 + 6x - 5 to find the value of y when x = -3:

y = (-3)^2 + 6(-3) - 5 = 9 - 18 - 5 = -14

Then the vertex is (-3, -14).</span>

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