I've a homework problem that I'm struggling to solve.
I need to find the value of $\alpha$ and $\beta$ that will make the function $$g(x) = 2\sqrt{x-\alpha} + \beta$$ pass through $(3, -2), (4,0), (7,2)$.
The best I got was try to convert the function to a straight line form by have $\alpha=x-x^2$ so that we left with $g(x) = 2x+\beta$ but I can't figure out what $\beta$ should be?
Maybe my assumption that the graph should be a straight line is wrong?
Thank you!
Yes, your assumption is not correct. The function $g(x) = 2\sqrt{x-\alpha} + \beta$ is not a straight line; it's a transformation of the square root function, which is not linear.
Let's use the three points you're given to create a system of equations that we can solve for the unknowns $\alpha$ and $\beta$.
We know that the function $g(x)$ passes through the points $(3,-2)$, $(4,0)$, and $(7,2)$. This means that when $x=3$, $g(x) = -2$; when $x=4$, $g(x) = 0$; and when $x=7$, $g(x) = 2$.
We can convert these into three equations:
Subtracting the second equation from the first gives: $\sqrt{3-\alpha} = -\sqrt{4-\alpha} - 1$
Squaring both sides to get rid of the square roots gives: $3 - \alpha = 1 + 2\sqrt{4-\alpha} + 4 - \alpha$ which simplifies to: $2\sqrt{4-\alpha} = \alpha - 2$
Again, squaring both sides to get rid of the square roots gives: $4(4-\alpha) = (\alpha - 2)^2$ which simplifies to: $16 - 4\alpha = \alpha^2 - 4\alpha + 4$ which simplifies to: $\alpha^2 = 16$
So, $\alpha = ±4$. But if we consider the domain of the square root function in $g(x)$, $\alpha$ must be less than or equal to the smallest $x$ value given (which is 3), so $\alpha = -4$.
Now, we can substitute $\alpha = -4$ into the second equation to find $\beta$: $2\sqrt{4 - (-4)} + \beta = 0$ $2\sqrt{8} + \beta = 0$ $\beta = -2\sqrt{8} = -4\sqrt{2}$
So, $\alpha = -4$ and $\beta = -4\sqrt{2}$ make the function $g(x) = 2\sqrt{x-\alpha} + \beta$ pass through the points $(3,-2)$, $(4,0)$, and $(7,2)$.
Your assumption that the graph is a line is incorrect. The graph of $g(x) = 2\sqrt{x - \alpha} + \beta$ is the upper half of a parabola that opens to the right and has endpoint $(\alpha, \beta)$. Since the graph passes through the points $(3, -2)$, $(4, 0)$, and $(7, 2)$, \begin{align*} g(3) & = 2\sqrt{3 - \alpha} + \beta = -2 \tag{1}\\ g(4) & = 2\sqrt{4 - \alpha} + \beta = 0 \tag{2}\\ g(7) & = 2\sqrt{7 - \alpha} + \beta = 2 \tag{3} \end{align*} Subtracting equation $1$ from equation $2$ yields $$2\sqrt{4 - \alpha} - 2\sqrt{3 - \alpha} = 2$$ Simplifying yields \begin{align*} \sqrt{4 - \alpha} - \sqrt{3 - \alpha} & = 1\\ \sqrt{4 - \alpha} & = 1 + \sqrt{3 - \alpha} \end{align*} Squaring both sides of the equation and solving for $\alpha$ yields \begin{align*} 4 - \alpha & = 1 + 2\sqrt{3 - \alpha} + 3 - \alpha\\ 0 & = 2\sqrt{3 - \alpha}\\ 0 & = \sqrt{3 - \alpha}\\ 0 & = 3 - \alpha\\ \alpha & = 3 \end{align*} Substituting $3$ for $\alpha$ in equation $1$ yields \begin{align*} 2\sqrt{3 - 3} + \beta & = -2\\ 2\sqrt{0} + \beta & = -2\\ 2 \cdot 0 + \beta & = -2\\ 0 + \beta & = -2\\ \beta & = -2 \end{align*} Hence, $g(x) = 2\sqrt{x - 3} - 2$.
Check: If $g(x) = 2\sqrt{x - 3} - 2$, then \begin{align*} g(3) & = 2\sqrt{3 - 3} - 2 = 2\sqrt{0} - 2 = 2 \cdot 0 - 2 = 0 - 2 = -2\\ g(4) & = 2\sqrt{4 - 3} - 2 = 2\sqrt{1} - 2 = 2 \cdot 1 - 2 = 2 - 2 = 0\\ g(7) & = 2\sqrt{7 - 3} - 2 = 2\sqrt{4} - 2 = 2 \cdot 2 - 2 = 4 - 2 = 2 \end{align*} as required.
It may be helpful to get rid of the square-root at the start by re-arranging the curve equation into $ \ (y - \beta)^2 \ = \ 4·(x - \alpha) \ \ , \ $ which suggests the equation of a "horizontal" parabola. (The curve equation we seek is a "shifted" version of the "upper half" of this parabola.)
Using the given three points produces $$ (-2 - \beta)^2 \ \ = \ \ 4·(3 - \alpha) \ \ \ , \ \ \ (0 - \beta)^2 \ \ = \ \ 4·(4 - \alpha) \ \ \ , \ \ \ (2 - \beta)^2 \ \ = \ \ 4·(7 - \alpha) \ \ . $$ The middle equation gives us the useful relation $ \ \beta^2 \ = \ 16 - 4·\alpha \ \ . \ $ Subtracting the first equation from the third gives us $$ (4 \ - \ 4·\beta \ + \ \beta^2) \ - \ (4 \ + \ 4·\beta \ + \ \beta^2) \ \ = \ \ (28 \ - \ 4·\alpha) \ - \ (12 \ - \ 4·\alpha) $$ $$ \Rightarrow \ \ -8·\beta \ \ = \ \ 16 \ \ , $$ leading to the values of $ \ \alpha \ $ and $ \ \beta \ \ $ by which the "basic" square-root function is translated "to the right" and "upward" (since $ \ \beta \ $ turns out to be negative, that "shift" is actually "downward").
