To find the x-intercepts, we set the numerator equal to zero and solve.

$\displaystyle 0=3x^2-5$

$\displaystyle 5=3x^2$

$\displaystyle \frac{5}{3}=x^2$

$\displaystyle \sqrt{\frac{5}{3}}=\sqrt{x^2}$

$\displaystyle \sqrt{\frac{5}{3}}=x$

However, the square root of a number can be both positive and negative.

Therefore the roots will be $\displaystyle x=\sqrt{\frac{5}{3}}, -\sqrt{\frac{5}{3}}.$

Begin by recognizing that both sides of the equation have a root term of $\displaystyle 3$.

$\displaystyle \small 9^x=3^6$

$\displaystyle (3^2)^x=3^6$

Using the power rule, we can set the exponents equal to each other.

$\displaystyle 3^{(2*x)}=3^6$

$\displaystyle \small 2x=6$

$\displaystyle \small x=3$

The question gives us P (48,000) and asks us to find t (time). We can substitute for P and start to solve for t:

$\displaystyle 48,000 = 2000e^{2t}$

$\displaystyle 24 = e^{2t}$

Now we have to isolate t by taking the natural log of both sides:

$\displaystyle \ln 24 = \ln e^{2t}$

$\displaystyle \ln 24 = (2t)\ln e$

And since $\displaystyle \ln e = 1$, t can easily be isolated:

$\displaystyle \ln 24 = 2t$

$\displaystyle \frac{\ln24}{2} = t$

Note: $\displaystyle \frac{\ln 24}{2}$ does not equal $\displaystyle \ln 12$ . You have to perform the log operation first before dividing.

Begin by recognizing that both sides of the equation have the same root term, $\displaystyle 3$.

$\displaystyle \small 3^{2x}=81$

$\displaystyle \small 3^{2x}=9^2$

$\displaystyle 3^{2x}=(3^2)^2$

We can use the power rule to combine exponents.

$\displaystyle 3^{2x}=3^4$

Set the exponents equal to each other.

$\displaystyle 2x=4$

$\displaystyle \small x=2$

Pull an $\displaystyle x$ out of the left side of the equation.

$\displaystyle \rightarrow x(x^2 - 9) =0$

Use the difference of squares technique to factor the expression in parentheses.

$\displaystyle \rightarrow x(x+3)(x-3) =0$

Any number that causes one of the terms $\displaystyle x$, $\displaystyle x+3$, or $\displaystyle x-3$ to equal $\displaystyle 0$ is a solution to the equation. These are $\displaystyle 0$, $\displaystyle -3$, and $\displaystyle 3$, respectively.

To find the $\displaystyle y$-intercept, set $\displaystyle x=0$ in the equation and solve.

$\displaystyle y=\frac{x^3+2x+8}{x^2-36}$

$\displaystyle y=\frac{(0)^3+2(0)+8}{(0)^2-36}$

$\displaystyle y=\frac{(0)+(0)+8}{(0)-36}$

$\displaystyle y=\frac{8}{-36}$

$\displaystyle y=-\frac{2}{9}$

To find the $\displaystyle y$-intercept(s) of $\displaystyle y=\frac{x^2+9x+8}{x^2-44}$, set the $\displaystyle x$ value equal to zero and solve.

$\displaystyle y=\frac{x^2+9x+8}{x^2-44}$

$\displaystyle y=\frac{(0)^2+9(0)+8}{(0)^2-44}$

$\displaystyle y=\frac{(0)+(0)+8}{(0)-44}$

$\displaystyle y=\frac{8}{-44}$

$\displaystyle y=-\frac{2}{11}$