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3 years ago

9

Graphing Exponential Functions In Exercise, sketch the graph of the function.

1 answer:

denis23 [38]3 years ago

5 0

Answer:

The graph of the function $f(x) = (1/2)^2x + 4$ is attached as image below.

Analysis:

We have two intersection points in the graph

The function $f(x) = (1/2)^2x + 4$ intersects the y-axis at points (0,4)

The function $f(x) = (1/2)^2x + 4$ intersects the x-axis at points (-16,0)

f(x) is positive for x>-`16

f(x) is negative for x<-`16

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Consider the equation: 12x=13-x^212x=13−x 2 12, x, equals, 13, minus, x, squared 1) Rewrite the equation by completing the squar

Flauer [41]

Answer:

$(x + 6)^2 = 49$

$x = 1$ or $x = -13$

Step-by-step explanation:

Given

$12x = 13 - x^2$

Using Completing the Square

$12x = 13 - x^2$ ---- Add $x^2$ to both sides

$x^2 + 12x = 13 - x^2 + x^2$

$x^2 + 12x = 13$

Divide the coefficient of x by 2; then add the square to both sides

$x^2 + 12x + 6^2 = 13 + 6^2$

$x^2 + 12x + 36 = 13 + 36$

$x^2 + 12x + 36 = 49$

Factorize

$x^2 + 6x + 6x + 36 = 49$

$x(x + 6) + 6(x + 6) = 49$

$(x + 6)(x + 6) = 49$

$(x + 6)^2 = 49$

Hence, the equation is $(x + 6)^2 = 49$

Solving further

Take square root of both sides

$(x + 6) = \sqrt{49}$

$x + 6 = \±7$

$x = \±7- 6$

This implies that

$x = 7 - 6$ or $x = -7 -6$

$x = 1$ or $x = -13$

HEnce, the solutions are $x = 1$ or $x = -13$

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3 years ago