Q:

Graph f (x) = 1.6x-2 +1describe the domain and rangedescribe x- and y- intercepts

Accepted Solution

A:
Answer:We need to find the domain, range, x-intercept and y-intercept of the following function:[tex]f(x) = 1.6x^{-2} + 1[/tex] β‡’ [tex]f(x)=\frac{1.6}{x^{2} }+1[/tex]To find the y-intercept, we have to make 'x=0'[tex]f(x) = \frac{1.6}{x^{2} } + 1[/tex] β‡’ [tex]f(x) = \frac{1.6}{0} Β + 1[/tex]. Given that divisions by zero are not possible, we conclude that there's no y-intercept. In other words, the function does not cross the y-axis, To find the x-intercept, we have to make 'y=0'[tex]f(x) = \frac{1.6}{x^{2} } + 1[/tex] Β β‡’ [tex]\frac{1.6}{x^{2} } + 1 = 0[/tex]β‡’ [tex]x^{2} = -1.6[/tex]Given that we cannot take the square rooth of a negative number, we can conclude that there's no x-intercept. In other words, the function does not cross the x-axis.The domain is all the possible values that the independent variable 'x' can take. Given that we can not divide by zero, the domain is all real numbers except zero. In set notation: ℝ - {0}.The Range is all the possible values that the dependent variable 'y' can take. Solving the expression for 'x' we have:[tex]\frac{1.6}{x^{2} } + 1 = y[/tex] Β β‡’ [tex]\frac{1.6}{x^{2} }= y-1[/tex]β‡’ [tex]\sqrt{(\frac{1.6}{y-1 })}= x[/tex] Given that square roots can not be negative, and the denominator can't be equal to zero, the range is y>1. In set notation: Range: (1, +∞)