How is the parabolic equation calculated

Find the parabolic equation from two points and one parameter

Often the equation of a parabola is to be determined for which two points are known and one of the parameters $ a $, $ b $ or $ c $ of the general form $ f (x) = ax ^ 2 + bx + c $. The third piece of information is often found hidden as a "shifted normal parabola", sometimes also opened downwards. On this page you will learn how to solve these and similar tasks.

Intuition

In the following graphic you can move the red points. You can change the stretching factor (opening factor) $ a $ using the slider. If you have not yet discussed the stretching factor in class: for $ a = 1 $ you get a normal parabola that opens upwards, for $ a = -1 $ you get a normal parabola that opens downwards.

As long as the points do not have the same abscissa ($ x $ coordinate), a function graph is created. For $ a \ not = 0 $ you get a parabola, otherwise a straight line.

Calculate the functional equation for a given stretch factor

The prerequisite is that you can solve simple systems of linear equations using the addition and subtraction method.

The most common case is the shifted normal parabola, i.e. $ a = 1 $.

example 1: We are looking for the equation of a shifted normal parabola, which is defined by the points $ A (\ color {# f00} {- 1} | \ color {# 1a1} {6}) $ and $ B (\ color {# a61} {3 } | \ color {# 18f} {- 1}) $ goes.

solution: Because of $ a = 1 $, a displaced normal parabola has an equation of the type $ f (x) = x ^ 2 + bx + c $. The coordinates of the points have to "fulfill the equation", that is to say they have to give a true statement when inserted. This leads to the following conditions:

$ \ begin {alignat *} {6} & f (\ color {# f00} {- 1}) = \ color {# 1a1} {6} \ quad && \ quad & (\ color {# f00} {- 1} ) ^ 2 & \, + \, & b \ cdot (\ color {# f00} {- 1}) & \, + \, & c & \, = \, & \ color {# 1a1} {6} \ & \ quad && \ text {I} \ quad & 1 & \, - \, & b & \, + \, & c & \, = \, & 6 \ & f (\ color {# a61} {3}) = \ color {# 18f} { -1} \ quad && \ quad & \ color {# a61} {3} ^ 2 & \, + \, & b \ cdot \ color {# a61} {3} & \, + \, & c & \, = \, & \ color {# 18f} {- 1} \ & \ quad && \ text {II} \ quad & 9 & \, + \, & 3b & \, + \, & c & \, = \, & - 1 \ end {alignat *} $

With a little practice, you will immediately write down the final equations I and II without the intermediate step of extensive filling in. The result is a system of equations with two unknowns that is easiest to solve using the subtraction method, since $ c $ is omitted in this way. It is up to you whether you bring the numbers 1 or 9 to the other side. It is not necessary for the manual procedure, but it is clearer.

$ \ begin {alignat *} {6} & \ text {I} \ quad & 1 & \, - \, & b & \, + \, & c & \, = \, & 6 \ qquad & | -1 \ & \ text {II } \ quad & 9 & \, + \, & 3b & \, + \, & c & \, = \, & - 1 \ qquad & | -9 \ \ & \ text {I} _a \ quad && \, \, & - b & \, + \, & c & \, = \, & 5 \ qquad & \ & \ text {II} _a \ quad && \, \, & 3b & \, + \, & c & \, = \, & - 10 \ qquad & \ \ & \ text {II} _a- \ text {I} _a \ quad && \, \, & 4b & \, \, && \, = \, & - 15 \ qquad & |: 4 \ & \ quad && \, \, & b & \, \, && \, = \, & - 3 {,} 75 \ qquad & \ \ & b \ text {in I} _a \ quad & - & \, (- & 3 {, } 75) & \, + \, & c & \, = \, & 5 \ qquad & \ & \ quad && \, \, & 3 {,} 75 & \, + \, & c & \, = \, & 5 \ qquad & | -3 {,} 75 \ & \ quad && \, \, && \, \, & c & \, = \, & 1 {,} 25 \ qquad & \ \ end {alignat *} $

The parabola we are looking for has the equation $ f (x) = x ^ 2-3 {,} 75x + 1 {,} 25 $.

Example 2: A parabola is stretched with the factor $ \ color {# 18f} {2} $ and opened downwards. It goes through the points $ A (-3 | -9) $ and $ B \ left (2 \ big | \ frac {23} {3} \ right) $. Find your equation.

solution: Since the parabola is open downwards, $ a = \ color {# f00} {-} \ color {# 18f} {2} $. We put in $ f (x) = - 2x ^ 2 + bx + c $, immediately write down the finished equations and subtract them without simplifying beforehand:

$ \ begin {alignat *} {6} & f (-3) = - 9 \ quad && \ text {I} \ quad & -18 & \, - \, & 3b & \, + \, & c & \, = \, & - 9 \ qquad & \ & f (2) = \ tfrac {23} {3} \ quad && \ text {II} \ quad & -8 & \, + \, & 2b & \, + \, & c & \, = \, & \ tfrac {23} {3} \ qquad & \ \ & \ quad && \ text {II} - \ text {I} \ quad & 10 & \, + \, & 5b & \, \, && \, = \, & \ tfrac {50} {3} \ qquad & | -10 \ & \ quad && \ quad && \, \, & 5b & \, \, && \, = \, & \ tfrac {20} {3} \ qquad & |: 5 \ & \ quad && \ quad && \, \, & b & \, \, && \, = \, & \ tfrac {4} {3} \ qquad & \ \ & \ quad && b \ text { in I} \ quad & -18 & \, - \, & 3 \ cdot \ tfrac 43 & \, + \, & c & \, = \, & - 9 \ qquad & | + 18 + 3 \ cdot \ tfrac 43 \ & \ quad && \ quad && \, \, && \, \, & c & \, = \, & 13 \ qquad & \ \ end {alignat *} $

The function equation is $ f (x) = - 2x ^ 2 + \ tfrac 43x + 13 $.

The y-axis intercept is given

The parameter $ c $ is the $ y $ axis section and can either be given directly (intersects the $ y $ axis at ...) or indirectly as a further point $ P (0 | c) $.

Example 3: A parabola cuts the $ y $ -axis at $ \ color {# b1f} {4} $ and goes through the point $ A (\ color {# a61} {2} | \ color {# 18f} {6}) $. In addition, a zero with $ x = \ color {# f00} {- 1} $ is known. What is their equation called?

solution: The intersection with the $ y $ axis provides the parameter $ c = \ color {# b1f} {4} $ and the zero a second point $ B (\ color {# f00} {- 1} | \ color {# 1a1} {0}) $. We therefore start from the equation $ f (x) = ax ^ 2 + bx + \ color {# b1f} {4} $ and insert the coordinates of both points:

$ \ begin {alignat *} {6} & f (\ color {# a61} {2}) = \ color {# 18f} {6} \ quad && \ text {I} \ quad & 4a & \, + \, & 2b & \ , + \, & 4 & \, = \, & 6 \ & f (\ color {# f00} {- 1}) = \ color {# 1a1} {0} \ quad && \ text {II} \ quad & a & \, - \, & b & \, + \, & 4 & \, = \, & 0 \ \ end {alignat *} $

Subtracting the equations is now unsuccessful because $ c $ is already known. Instead, we eliminate $ b $ and, for this purpose, multiply equation II by 2:

$ \ begin {alignat *} {6} & \ text {I} \ quad & 4a & \, + \, & 2b & \, + \, & 4 & \, = \, & 6 \ qquad & \ & \ text {II} \ cdot 2 \ quad & 2a & \, - \, & 2b & \, + \, & 8 & \, = \, & 0 \ qquad & \ \ & \ text {II} \ cdot 2+ \ text {I} \ quad & 6a & \, \ , && \, + \, & 12 & \, = \, & 6 \ qquad & | -12 \ & \ quad & 6a & \, \, && \, \, && \, = \, & - 6 \ qquad & |: 6 \ & \ quad & a & \, \, && \, \, && \, = \, & - 1 \ qquad & \ \ & a \ text {in II} \ quad & -1 & \, - \, & b & \ , + \, & 4 & \, = \, & 0 \ qquad & | + 1-4 \ & \ quad && \, - \, & b & \, \, && \, = \, & - 3 \ qquad & | :( -1) \ & \ quad && \, \, & b & \, \, && \, = \, & 3 \ qquad & \ \ end {alignat *} $

The equation is $ f (x) = - x ^ 2 + 3x + 4 $. Because of $ a = -1 $, the parabola has the shape of a normal parabola that opens downwards.

More options

If the parameter $ b $ is given next to two dots, proceed in a similar way to example 2.

If both zeros are given (i.e. the points of intersection with the $ x $ axis), you can proceed as here or use the zero equation (linear factor form). The zero point approach is faster, especially with a given $ a $ or $ c $, but is by no means dealt with in all schools.

Exercises

Last update: 02.12.2015; © Ina de Brabandt

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