Exercise 12 Page 212

Before we try to write the equation of a specific parabola that passes through some given points, let's make sure we have enough points. We will need at least three. (1,0),(- 2,9),(- 4,5) To use the given points, we need to substitute their ( x, y) coordinate pairs into the standard form of a quadratic equation. y=a x^2+b x+c, a≠ 0 Doing so will create a system of equations that we can solve for the values of a, b, and c. Let's start with (1,0). We just wrote our first equation! Now let's do the same thing for (- 2,9). To find our third and last equation, we will use (- 4,5). We now have a system of three equations. a+b+c=0 & (I) 4a-2b+c=9 & (II) 16a-4b+c=5 & (III) Let's solve this system using the Elimination Method. We will start by subtracting Equation (I) from Equation (II) to eliminate the c-variable. Now let's subtract Equation (I) from Equation (III) to eliminate the c-variable once more. Now, neither Equation (II) nor Equation (III) includes the c-variable. These equations form a system with only two variables, a and b. Let's solve this system by using the Elimination Method again. Since neither variable has the same or opposite coefficients, we will need to multiply one equation by a number first. We found our first value, allowing us to form a partial equation. y=ax^2+(- 4)x+c ⇔ y=ax^2-4x+c Let's substitute - 4 for b in Equation (III) to find the value of a. With our second value, we can continue forming the partial equation. y=(- 1)x^2+(-4)x+c ⇔ y=- x^2-4x+c Finally, to find the value of c, we will substitute a=- 1 and b=- 4 into Equation (I). Now that we have all three values, we can complete the standard form equation of the parabola that passes through the given points. y=(-1)x^2+(-4)x+5 ⇔ y=- x^2-4x+5 To help visualize this graph, we have plotted the given points and sketched the curve below.