Question

given the following table:

x	f(x)
4	0
9	1
16	2

determine the values for a, h, and k using regression.

a =

h =

k =

Explanation:

Step1: Identify model form

We assume the model is $f(x) = a\sqrt{x - h} + k$. First, linearize it by letting $y = f(x)$, $z = \sqrt{x}$. Then the model becomes $y = a z + (k - a\sqrt{h})$, which is linear in $z$.

Step2: Calculate $z = \sqrt{x}$ values

For $x=1$: $z_1=\sqrt{1}=1$
For $x=4$: $z_2=\sqrt{4}=2$
For $x=9$: $z_3=\sqrt{9}=3$
For $x=16$: $z_4=\sqrt{16}=4$

Step3: Compute regression for $y=az+b$

We use linear regression formula for $a = \frac{n\sum z_iy_i - \sum z_i\sum y_i}{n\sum z_i^2 - (\sum z_i)^2}$, where $b = k - a\sqrt{h}$
First calculate sums:
$\sum z_i = 1+2+3+4=10$
$\sum y_i = -1+0+1+2=2$
$\sum z_iy_i = (1)(-1)+(2)(0)+(3)(1)+(4)(2) = -1+0+3+8=10$
$\sum z_i^2 = 1^2+2^2+3^2+4^2=1+4+9+16=30$
$n=4$
$a = \frac{4(10) - (10)(2)}{4(30) - (10)^2} = \frac{40-20}{120-100}=\frac{20}{20}=1$

Step4: Calculate $b$

$b = \frac{\sum y_i - a\sum z_i}{n} = \frac{2 - 1(10)}{4}=\frac{2-10}{4}=-2$

Step5: Solve for $h$ and $k$

Since $b = k - a\sqrt{h}$, and $a=1$, we have $k - \sqrt{h} = -2$. Notice from the table, when $x=4$, $f(x)=0$:
$0 = 1\sqrt{4 - h} + k \implies \sqrt{4-h} = -k$
Substitute $k = \sqrt{h} - 2$ into this:
$\sqrt{4-h} = -(\sqrt{h}-2) = 2 - \sqrt{h}$
Square both sides:
$4 - h = 4 -4\sqrt{h} + h$
$0 = 2h -4\sqrt{h}$
$2\sqrt{h}(\sqrt{h}-2)=0$
$\sqrt{h}=2$ (since $\sqrt{h}=0$ would not fit the data), so $h=4$
Then $k = \sqrt{4} -2 = 2-2=0$

Answer:

$a = 1$
$h = 4$
$k = 0$