Question

Find the quadratic polynomial whose graph goes through the points (−2,10),(−2,10), (0,6),(0,6), and (1,10)(1,10).

Answer 1

Answer on Question #61597–Math –Linear Algebra

Question

Find the quadratic polynomial whose graph goes through the points $(-2,10)$, $(0,6)$ and $(1,10)$.

Solution

Consider the quadratic polynomial

$$y(x) = ax^2 + bx + c$$

and substitute the given coordinates of points:

$y(-2) = 10$ then $a \cdot (-2)^2 + b \cdot (-2) + c = 10$, that is, $4a - 2b + c = 10$;

$y(0) = 6$ then $a \cdot 0^2 + b \cdot 0 + c = 6$, so $c = 6$;

$y(1) = 10$ then $a \cdot 1^2 + b \cdot 1 + c = 10$, that is, $a + b + c = 10$.

So we get the system of three equations which we can be solved using elimination:

$$\begin{array}{c} c = 6, \\ 4a - 2b + c = 10, \\ a + b + c = 10. \end{array}$$

Since $c = 6$, we can substitute it into the other two equations to get:

$$4a - 2b + 6 = 10 \text{ and } a + b + 6 = 10.$$

Simplifying to

$$\begin{array}{l} 4a - 2b = 4, \quad (2) \\ a + b = 4. \quad (3) \end{array}$$

Dividing (2) by 2

$$2a - b = 2.$$

It follows from (3) that

$$b = 4 - a.$$

Using (5) and substituting for $b$ into (4) one gets

$$\begin{array}{l} 2a - (4 - a) = 2; \\ 2a - 4 + a = 2, \\ 3a = 6, \\ a = 2. \end{array}$$

Then $b = 4 - a = 4 - 2 = 2$.

Thus, $a = 2$, $b = 2$ and $c = 6$.

Substituting those values into (1) one gets the general equation of the quadratic polynomial that passes through the three given points:

$$y(x) = 2x^2 + 2x + 6.$$

Answer: $y(x) = 2x^2 + 2x + 6$.

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