Exercise 1 Page 212

Let's start by recalling the standard form of a quadratic function. To find the equation of a parabola that includes the given points, we will substitute their coordinates into the above equation and simplify. With the resulting equations, we will write a system of equations. Then, we will solve it to find the coefficients $a,$ $b,$ and $c.$

$y=a{x}^2+bx+c$
Point	Substitute	Simplify
$(1,0)$	$0=a(1{)}^2+b(1)+c$	$a+b+c=0$
$(2,\text{-}3)$	$\text{-}3=a(2{)}^2+b(2)+c$	$4a+2b+c=\text{-}3$
$(3,\text{-}10)$	$\text{-}10=a(3{)}^2+b(3)+c$	$9a+3b+c=\text{-}10$

We can now write a system of three equations. Let's solve this system using the Elimination Method. We will start by subtracting Equation (I) from Equation (II) and Equation (III) to eliminate $c.$ Now, neither Equation (II) nor Equation (III) includes $c.$ These equations form a system in terms of only $a$ and $b.$ Let's solve this system by using the Elimination Method again. Since neither variable has the same or opposite coefficients, we need to multiply or divide one equation by a number first. We found our first value, allowing us to write a partial equation. Let's substitute $\text{-}2$ for $a$ in Equation (II) to find the value of $b.$ With our second value, we can continue forming the partial equation. Finally, to find the value of $c,$ we will substitute $a=\text{-}2$ and $b=3$ into Equation (I). Now that we have all three values, we can complete the equation of the parabola that passes through the given points. To help visualize this situation, we can plot the given points and sketch the curve.