Finding the Equation of a Parabola Lesson

In some cases, we may be asked to find the equation of a parabola given three points on the parabola. To accomplish this task, we set up and solve a system of equations with three equations and three unknowns. Let's look at an example.
Example #1: Find the equation of the parabola that goes through the given points. $$y=ax^2 + bx + c$$ $$(0, 4)$$ $$(2, -2)$$ $$(3, 10)$$ We need to find a, b, and c. These are the three unknowns. Let's set up a system of equations. For each equation, we can plug in for x and y:
Plug in a 0 for x and a 4 for y: $$y = ax^2 + bx + c$$ $$4 = a(0)^2 + b(0) + c$$ $$4=c$$ Plug in a 2 for x and a -2 for y: $$y = ax^2 + bx + c$$ $$-2 = a(2)^2 + b(2) + c$$ $$-2=4a + 2b + c$$ Plug in a 3 for x and a 10 for y: $$y = ax^2 + bx + c$$ $$10 = a(3)^2 + b(3) + c$$ $$10=9a + 3b + c$$ Now that we have our system, let's solve for a, b, and c: $$1) \, c = 4$$ $$2) \, 4a + 2b + c = -2$$ $$3) \, 9a + 3b + c = 10$$ Since c = 4, we can plug a 4 in for c in equation #2 and equation #3. Let's start with equation #2. $$2) \, 4a + 2b + 4 = -2$$ Subtract 4 away from each side: $$2) \, 4a + 2b = -6$$ To make the equation simpler, we can divide both sides by 2: $$2) \, 2a + b = -3$$ Let's do the same thing for equation #3. $$3) \, 9a + 3b + 4 = 10$$ Subtract 4 away from each side: $$3) \, 9a + 3b = 6$$ Again to make the equation simpler, we can divide both sides by 3: $$3) \, 3a + b = 2$$ Now we can solve a linear system in two variables (a and b). $$2) \, 2a + b = -3$$ $$3) \, 3a + b = 2$$ Multiply equation #2 by -1, this will give us opposite coefficients on b. $$2) \, {-}2a - b = 3$$ Add equations #2 and #3: $$-2a - b = 3$$ $$\underline{+3a + b = 2}$$ $$a = 5$$ In order to find b, we can plug in for a using equation #3. $$3(5) + b = 2$$ $$15 + b = 2$$ $$b = -13$$ After solving our system, we end up with: $$a=5, b=-13, c=4$$ This gives us the following parabola: $$y = ax^2 + bx + c$$ $$y=5x^2 - 13x + 4$$ Desmos Link for More Detail Additionally, we may see this type of problem given with a sideways parabola, which is also known as a horizontal parabola. This type of parabola will be studied in more detail later on in the course when we get to conic sections. For now, we should know that the horizontal parabola is not a function as the "x" and "y" have swapped roles. The horizontal parabola has the following equation: $$x = ay^2 + by + c$$ At this point, we can just follow the same procedure that we did above. Let's look at an example.
Example #2: Find the equation of the parabola that goes through the given points. $$x = ay^2 + by + c$$ $$(2, 0)$$ $$(6, 1)$$ $$(4, -1)$$ We need to find a, b, and c. These are the three unknowns. Let's set up a system of equations. For each equation, we can plug in for x and y:
Plug in a 2 for x and a 0 for y: $$x = ay^2 + by + c$$ $$2 = a(0)^2 + b(0) + c$$ $$2 = c$$ Plug in a 6 for x and a 1 for y: $$x = ay^2 + by + c$$ $$6 = a(1)^2 + b(1) + c$$ $$6 = a + b + c$$ Plug in a 4 for x and a (-1) for y: $$x = ay^2 + by + c$$ $$4 = a(-1)^2 + b(-1) + c$$ $$4 = a - b + c$$ Now that we have our system, let's solve for a, b, and c: $$1) \, c = 2$$ $$2) \, a + b + c = 6$$ $$3) \, a - b + c = 4$$ Since c = 2, we can plug a 2 in for c in equation #2 and equation #3. Let's start with equation #2. $$2) \, a + b + 2 = 6$$ Subtract 2 away from each side: $$2) \, a + b = 4$$ Let's do the same thing for equation #3. $$3) \, a - b + 2 = 4$$ Subtract 2 away from each side: $$3) \, a - b = 2$$ Now we can solve a linear system in two variables (a and b). $$2) \, a + b = 4$$ $$3) \, a - b = 2$$ Add equations #2 and #3: $$+a + b = 4$$ $$\underline{+a - b = 2}$$ $$2a = 6$$ $$a = 3$$ In order to find b, we can plug in for a using equation #2. $$3 + b = 4$$ $$b = 1$$ After solving our system, we end up with: $$a=3, b=1, c=2$$ This gives us the following parabola: $$x = 3y^2 + y + 2$$ Desmos Link for More Detail

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