Given y = ax2 + bx and a table of x and y values, estimate the values of a and b.

The given equation is $y = ax^2 + bx$.
To make it linear, we can divide the entire equation by $x$ (assuming $x \neq 0$):

$$\frac{y}{x} = \frac{ax^2}{x} + \frac{bx}{x}$$ $$\frac{y}{x} = ax + b$$

Now, let $Y = \frac{y}{x}$ and $X = x$. The equation becomes:

$$Y = aX + b$$

This is in the form of a straight line $Y = mX + c$, where the gradient $m = a$ and the Y-intercept $c = b$.

The original table is:

$x$	1	2	3	4	5	6
$y$	6	16	30	48	70	96

Now, we calculate $Y = \frac{y}{x}$ for each pair:

• For $x=1, y=6$: $Y = \frac{6}{1} = 6$
• For $x=2, y=16$: $Y = \frac{16}{2} = 8$
• For $x=3, y=30$: $Y = \frac{30}{3} = 10$
• For $x=4, y=48$: $Y = \frac{48}{4} = 12$
• For $x=5, y=70$: $Y = \frac{70}{5} = 14$
• For $x=6, y=96$: $Y = \frac{96}{6} = 16$

The new table for plotting the straight-line graph is:

$X = x$	$Y = \frac{y}{x}$
1	6
2	8
3	10
4	12
5	14
6	16

To draw the graph:

• Plot the points $(X, Y)$ from the table above on a graph paper.
• The horizontal axis (X-axis) should represent $x$.
• The vertical axis (Y-axis) should represent $\frac{y}{x}$.
• Draw a straight line that passes through all these points. (In this case, the points are perfectly collinear).

From the linear equation $Y = aX + b$:

• The value of $a$ is the gradient of the straight line.
• The value of $b$ is the Y-intercept (the value of $Y$ when $X=0$).

To find the gradient $a$:
We can use any two points from the transformed table, for example, $(X_1, Y_1) = (1, 6)$ and $(X_2, Y_2) = (6, 16)$.

$$a = \frac{Y_2 - Y_1}{X_2 - X_1} = \frac{16 - 6}{6 - 1} = \frac{10}{5} = 2$$

To find the Y-intercept $b$:
We can substitute the value of $a$ and any point $(X, Y)$ into the equation $Y = aX + b$. Let's use the point $(1, 6)$:

$$6 = (2)(1) + b$$ $$6 = 2 + b$$ $$b = 6 - 2$$ $$b = 4$$

Thus, the estimated values are $a=2$ and $b=4$.

The final answer is $a=2, b=4$.