I need to find the standard form

Question

It passes through (-2,7) (1,10) and (2,27)

Answer 1

Hello, Diana!

When trying to find the standard form of a quadratic equation from three points (or even two points), the best way to approach the problem is to set up a system of equations! Let's analyze Bradford T's answer a little more thoroughly.

Remember that the points given to you are essentially x and y-values. You can therefore plug them into the standard equation and simplify.

For example, for the point (-2,7), we plug in -2 for x and 7 for y and simplify.

y=ax²+ bx+c

7 = a(-2)² + b(-2) + c

7 = 4a -2b + c

We can repeat this for the other two points.

(1,10) -----> y= ax² + bx + c -----> 10 = a(1)² + b(1) + c ----> 10 = 1a + 1b + c

(2,27) -----> y= ax²+ bx + c ------> 27 = a(2)² + b(2) + c ----> 27 = 4a + 2b + c

Now we have the three equations.

7 = 4a -2b + c

10 = 1a + 1b + c

27 = 4a + 2b + c

Now, we should already be familiar with how to solve systems of equations with two variables. But how in the world do we solve a system of equations with three variables?

There are two options: to use a matrix (if your class has covered that concept, this can be the easiest method) or to use regular substitution or elimination! I prefer to use substitution or elimination.

So how do we use substitution or elimination for three variables? Simple! We deal with only two equations at a time!

I will label the equations D, F, G to explain this.

Equation D: 7 = 4a -2b + c

Equation F: 10 = 1a + 1b + c

Equation G: 27 = 4a + 2b + c

We will eliminate a chosen variable from equations D and F. Then, we will eliminate *the same* variable from equations F and G. From the two new equations that we produced, we will once again apply substitution or elimination.

I will choose to eliminate c from equations D, F, and G since the coefficient is 1 in all equations. It does not matter which variable you choose to eliminate first! But try to examine all three equations and see which variable has smaller coefficients or coefficients already with opposite signs so that you don't have to handle larger numbers and the opposite signs make it easier to eliminate.

Now that we have a plan, let's get to it!

Equation D: 7 = 4a -2b + c

Equation F: (-1) 10 = 1a + 1b + c (-1) **I am multiplying all of equation F by -1 to cancel the variable c

7 = 4a - 2b + c

-10 = -1a -1b - c

-------------------------------------------------

Equation H: -3 = 3a -3b

Now I need to eliminate "c" from equation F and G.

Equation F; 10 = 1a + 1b + c

Equation G: 27 = 4a + 2b + c

(-1) 10 = 1a + 1b + c * (-1) (Again, I am multiplying Equation F by -1 on both sides to cancel c)

-10 = -1a -1b - c

27 = 4a + 2b + c

---------------------------

17 = 3a + b Equation J

Now, I have two new equations that I got after I eliminated "c" from all three equations.

Equation H: -3 = 3a -3b

Equation J: 17 = 3a + b

I must now apply substitution and elimination to Equation H and Equation J. Looking at the equations, there is a 3a in both equations. It would be very easy to multiply just one of them by a -1 to cancel the "3a" and eliminate.

(-1) -3 = 3a - 3b (-1)

3 = -3a + 3b

17 = 3a + b

--------------------------

20 = 4b

b = 5

Now that we've found b, we can plug b back into Equation H or J. I will plug it into J. 17 = 3a + 5

12 = 3a a= 4

Excellent! Now, that we've found both b and a, we can plug them both back into one of the original equations (D, F, or G) to find "c".

Equation D 7 = 4a -2b + c ----> 7 = 4(4) -2(5) + c ----> 7 = 16 - 10 + c 7 = 6 + c c = 1

Now we have finally found a, b, and c.

a = 4, b = 5, c = 1

We can plug these values into the equation for standard form.

y = 4x² + 5x + 1

In some classes, this is the "standard" form for the parabola. However, in higher-level math courses, this is not the end of the problem! So, at this point, please consider the class you are taking and whether or not your teacher has said this is the standard form that he or she is looking for.

The other "standard form" of a parabola is (x-h)²=4p(y-k) where (h,k) is the vertex. If your class has discussed conics, finding the focus, or finding the directrix of a parabola, then you should use this form.

We actually do not have to complete the square to find this form, although that is one way to do it! An easier way to do this is to use -b/2a to find the axis of symmetry and the x-value of the vertex and then plug the x-value back in to find y. Then, you would have to know that 4p = 1/a (where p will help you find the focus and the directrix).

y = 4x² + 5x + 1

our b value is 5 and a = 4. -b/2a ---> -5/2(4) --> -5/8

This is the x-value of the vertex and also the axis of symmetry.

Now we can plug this value back in for x to find y.

y = 4(-5/8)²+5(-5/8) + 1

= 4(25/64) -25/8 + 1

= 100/64 -25/8 + 1

= 25/16- 25/8 + 1 (100/64 was simplified by dividing by 4)

= 25/16- 50/16 + 16/16 (I converted everything into 16ths)

= -9/16

The vertex is (-5/8, -9/16) which gives us (h,k) where h = -5/8 and k = -9/16.

Now we just need to find 4p, which is 1/a. We know that our a value was 4, so this is just 1/4.

(x-h)²=4p(y-k)

(x+5/8)²= 1/4(y+9/16)

We are now finally done with our problem!

If you have any more questions, feel free to contact me to explain more Algebra 2 concepts!

Answer 2

y = ax²+bx+c

So we need to solve for a,b and c, which means we need 3 equations to solve for the 3 unknowns

Substituting the (-2,7),(1,10) and (2,27):

4a -2b+c = 7

a+b+c = 10

4a+4b+c = 27

Solving for the 3 unknowns:

a = 4

b = 5

c = 1

y = 4x²+5x+1

But the problem was asking for the standard form

y = A(x-h)² + k where (k,h) is the vertex of the parabola.

To convert to standard form, we need to complete the square

y = 4(x + 5/8)² +1 -25/16

y = 4(x+5/8)²-9/16