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4 years ago

Explain how to graph this exponential function.

Mathematics

1 answer:

gladu [14]4 years ago

8 0

Answer:

The domain is -∞ < x < ∞

The range is y > -1

The x-intercept is 0

The y-intercept is 0

Step-by-step explanation:

∵ f(x) = 0.4^x - 1

∴ The domain is all real numbers -∞ < x < ∞

∴ The range of it is y > -1

Note: If the range of f(x) = b^x , where b > 0 and b ≠ 1 is y > 0 ,

If f(x) is shifted down 1 unit (f(x) - 1) , then the range will be y > -1

∵ f(x) = 0 ⇒ 0.4^x - 1 = 0

∴ 0.4^x = 1 ⇒ ∴ x = 0

∴ The x-intercept is 0

∵ f(0) = 0.4^0 - 1 = 1 - 1 = 0

∴ The y-intercept is 0

∴ The graph of f(x) = 0.4^x - 1 passing through the origin point

Table:

x ⇒ f(x)

-2 ⇒ 5.25

-1 ⇒ 1.5

0 ⇒ 0

1 ⇒ -0.6

2 ⇒ -0.84

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cluponka [151]

Answer:

See explanation

Step-by-step explanation:

Prove that

$1^2+2^2+3^3+...+n^2=\dfrac{1}{6}n(n+1)(2n+1)$

1. When $n=1,$ we have

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2. Assume that for all $k$ following equality is true

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3. Prove that for $k+1$ the following equality is true too.

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$1^2+2^2+3^2+...+(k+1)^2=\\ \\=(1^2+2^2+3^3+...+k^2)+(k+1)^2=\\ \\=\dfrac{1}{6}k(k+1)(2k+1)+(k+1)^2=\\ \\=(k+1)\left(\dfrac{1}{6}k(2k+1)+k+1\right)=\\ \\=(k+1)\dfrac{2k^2+k+6k+6}{6}=\\ \\=(k+1)\dfrac{2k^2+7k+6}{6}=\\ \\=(k+1)\dfrac{2k^2+4k+3k+6}{6}=\\ \\=(k+1)\dfrac{2k(k+2)+3(k+2)}{6}=\\ \\=(k+1)\dfrac{(k+2)(2k+3)}{6}$

Consider right part:

$\dfrac{1}{6}(k+1)((k+1)+1)(2(k+1)+1)=\\ \\\dfrac{1}{6}(k+1)(k+2)(2k+3)$

We get the same left and right parts, so the equality is true for $k+1.$

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