Find the x-intercept of the function f(x) = a^x + 1 - 1.

1 answer:

4 0

I am not certain of what you wrote for the function but will assume that is likely exponential function or quadratic function.

First, (quadratic function) To solve this, you must know the quadratic formula. The x-intercept is value(s) that has the output value(y) of 0.

If the vertex of the quadratic function of (0,0), there is only one x-intercept. The number and value of the x-intercept depends on the slope and vertical displacement.

Second, (exponential function) note that there is no x-intercept. For instance, if a is 2, is there such value y that 2^y is 0? The smallest exponential value that is an integer is 1. Even broadening the limit to rational numbers, no such exponential value can have the result of 0. Therefore, in the basic form of exponential function, there is no x-intercept.

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What is the range of possible values for x?

Talja [164]

The minimum value for 2x is 0
the maximum value is achieved when A, D and C are collinear and the quadrilateral ABCD becomes an isosceles triangle ABC 
base AB = 52 and vertical angle 2x + 34° 

For the sine law 
(sin 2x)/22 = (sin ADB)/AB 
(sin 34°)/30 = (sin BDC)/BC 

is given that AB = BC, and sin ADC = sin BDC because they are supplementary, so from 
(sin ADC)/AB = (sin BDC)/BC 

it follows 
(sin 2x)/22 = (sin 34°)/30 

sin 2x = 22 (sin 34°)/30 

2x = asin(22 (sin 34°)/30) ≈ 24.2° 

x = 0.5 asin(22 (sin 34°)/30) ≈ 12.1° 

0 < x < 12.1°

7 0

2 years ago

The line passes through (9,8) and is perpendicular to x-2y=-16

pickupchik [31]

The equation of line perpendicular to x-2y=-16 passing through (9,8) is: y=-2x+26

Step-by-step explanation:

Given

$x-2y=-16\\Adding\ 2y\ on\ both\ sides\\x=2y-16\\Adding\ 16\ on\ both\ sides\\x+16=2y-16+16\\x+16=2y\\2y=x+16\\Dividing\ both\ sides\ by\ 2\\\frac{2y}{2}=\frac{x+16}{2}\\y=\frac{x}{2}+\frac{16}{2}\\y=\frac{1}{2}x+8\\$