Finding quadratic equation, given a graph?

[MathNoob94]

Hi, So I can't seem to figure out how to find the quadratic equation when given a graph the graph has the points (-2,2) and (0,1) and (1,-2.5)

Thank You!

 

[Deleted member 4993]

Hi, So I can't seem to figure out how to find the quadratic equation when given a graph the graph has the points (-2,2) and (0,1) and (1,-2.5)

Thank You!

Do you know how to solve simultaneous linear equations for 3 unknowns?

 

[Steven G]

Hi, So I can't seem to figure out how to find the quadratic equation when given a graph the graph has the points (-2,2) and (0,1) and (1,-2.5)

Thank You!

What have you tried?? Do you know the general equation of a quadratic equation?

 

[Ishuda]

Hi, So I can't seem to figure out how to find the quadratic equation when given a graph the graph has the points (-2,2) and (0,1) and (1,-2.5)

Thank You!

There are several ways to find the quadratic equation. Here is another which lets you find an equation but may be tedious to simplify: Suppose you want a polynomial going through some points, and just to make it easy but only as an example, we will use just one of your points (-2,2) and two others (0,4) and (4,32). Let's start with a 'point slope' type of form to go through the first point
f(x) = 2 + (x+2) g(x)
where g(x) can be any polynomial. Now the second point (0,4)
f(0) = 2 + (0+2) g(x) = 2 + 2 g(0) = 4
so if g(0) is 1 then f(0) will be 4 and we have
g(x) = 1 + (x-0) h(x) = 1 + x h(x)
or
f(x) = 2 + (x + 2) [1 + x h(x)]
and we are ready for the third point (4, 32).
f(4) = 2 + 6 * [1 + 4 h(4)] = 8 + 24 h(4) = 32
So, h(4) is equal to 1 or
h(x) = 1 + (x-4) k(x)
and
f(x) = 2 + (x + 2) {1 + x [1 + (x-4) k(x)]}
Now we notice that if k(x) is not zero, then we will have at least a cubic equation and if it is zero we will have a quadratic equation which is what we want. Therefore the quadratic equation which goes through the points (-2, 2), (0,4), and (4,32) is for k(x)=0 and we have
f(x) = 2 + (x + 2) {3 + x [1 + (x-4) 0]} = 2 + (x + 2) (1 + x )
or
f(x) = x² + 3 x + 4

 

[Steven G]

There are several ways to find the quadratic equation. Here is another which lets you find an equation but may be tedious to simplify: Suppose you want a polynomial going through some points, and just to make it easy but only as an example, we will use just one of your points (-2,2) and two others (0,4) and (4,32). Let's start with a 'point slope' type of form to go through the first point
f(x) = 2 + (x+2) g(x)
where g(x) can be any polynomial. Now the second point (0,4)
f(0) = 2 + (0+2) g(x) = 2 + 2 g(0) = 4
so if g(0) is 1 then f(0) will be 4 and we have
g(x) = 1 + (x-0) h(x) = 1 + x h(x)
or
f(x) = 2 + (x + 2) [1 + x h(x)]
and we are ready for the third point (4, 32).
f(4) = 2 + 6 * [1 + 4 h(4)] = 8 + 24 h(4) = 32
So, h(4) is equal to 1 or
h(x) = 1 + (x-4) k(x)
and
f(x) = 2 + (x + 2) {1 + x [1 + (x-4) k(x)]}
Now we notice that if k(x) is not zero, then we will have at least a cubic equation and if it is zero we will have a quadratic equation which is what we want. Therefore the quadratic equation which goes through the points (-2, 2), (0,4), and (4,32) is for k(x)=0 and we have
f(x) = 2 + (x + 2) {3 + x [1 + (x-4) 0]} = 2 + (x + 2) (1 + x )
or
f(x) = x² + 3 x + 4

Nice, thanks for showing us this.

 

[MathNoob94]

Thank You, after finding an example online it makes much more sense! I simply plug in the points and since one point is a y intercept it gives me c = 1 and from there I simply plug in and then use addition method to solve for each variable.

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