For: https://socratic.org/users/jeish-g The points with coordinates (-1,4) and (1,-8) lie on a curve which has the rule of the form y=aroot(3)(x)+b. Find the values of a and b.?

Jan 4, 2018

$a = - 6 , b = - 2$

Explanation:

We have two points for which we know the $x$ and $y$ coordinates. We can just plug these two points into the equation and solve for $a$ and $b$:
$- 8 = a \sqrt[3]{1} + b$

$4 = a \sqrt[3]{- 1} + b$

We can subtract these two equations from each other to cancel the $b$'s:
$- 8 - 4 = a \sqrt[3]{1} + b - \left(a \sqrt[3]{- 1} + b\right)$

$- 12 = a \cdot 1 + b - \left(- a + b\right)$

$- 12 = a + b + a - b$

$- 12 = 2 a$

$a = - \frac{12}{2} = - 6$

Now we can plug in $a = - 6$ and solve for $b$:
$4 = - 6 \sqrt[3]{- 1} + b$

$4 = 6 + b$

$b = 4 - 6 = - 2$

So, the final equation is $y = - 6 \sqrt[3]{x} - 2$

Jan 4, 2018

Reworkded:

$a = - 6$
$b = - 2$

Explanation:

$\textcolor{b r o w n}{\text{Formatting tip - write "color(white)("d")" hash y = a root(3)(x)+b hash }}$
$\textcolor{b r o w n}{\text{giving: ............."color(white)("ddddddddddddd}} y = a \sqrt[3]{x} + b$

Have a look at: https://socratic.org/help/symbols
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Set point 1 as ${P}_{1} \to \left({x}_{1} , {y}_{1}\right) = \left(- 1 , 4\right)$
Set point 2 as ${P}_{2} \to \left({x}_{2} , {y}_{2}\right) = \left(1 , - 8\right) \textcolor{red}{\leftarrow \text{ +8 corrected to -8}}$

Given:$\textcolor{w h i t e}{\text{d}} y = a \sqrt[3]{x} + b$

${P}_{1} \to + 4 = a \sqrt[3]{- 1} + b \text{ "...............Equation(1)" checked}$
P_2->color(white)(".")color(red)(-8)=aroot(3)(+1)+b" "...............Equation(2)" corrected"

$\textcolor{b l u e}{\text{Determine the value of } b}$

Note that:
(-1)xx(-1)xx(-1)color(white)("d")=color(white)("d")(-1)" so "root(3)(-1)=-1

Note that $\sqrt[3]{+ 1} = 1$

So we have:

color(white)("d")+4=-a+b" "....................Equation(1_a)
ul(color(white)("d")-8=+a+b)" "....................Equation(2_a)
color(green)(color(white)("d")-4=color(white)(".dd")0+2b) larr" "Eqn(1_a)+Eqn(2_a)

Divide both sides by $\textcolor{red}{2}$

$\textcolor{g r e e n}{- \frac{4}{\textcolor{red}{2}} = \frac{2 b}{\textcolor{red}{2}}}$

$- 2 = b \textcolor{w h i t e}{\text{d}}$ more formally presented as: $b = - 2$
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$\textcolor{b l u e}{\text{Determine the value of } a}$

Choosing $E q u a t i o n \left({2}_{a}\right)$ substitute for $b$ where $\textcolor{red}{b = - 2}$

$\textcolor{g r e e n}{- 8 = + a + \textcolor{red}{b} \textcolor{w h i t e}{\text{ddd") -> color(white)("ddd}} - 8 = a + \left(\textcolor{red}{- 2}\right)}$

$\textcolor{g r e e n}{\textcolor{w h i t e}{\text{dddddddddddddd")->color(white)("ddd}} - 6 = a}$
More formally $a = - 6$