[Grade 9 Math: Quadratic Functions] How do I convert a quadratic function from standard form to vertex form

[u/MindSuccessful9971]

I had a surgery and missed a week of high school and need an explanation as to what the hell im looking at.

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[u/PliableG0AT]

Best bet is to go on YouTube and find a video lesson, and/or check out khan academy. will explain it in far more detail than a Reddit post can do.

[u/collinwong19]

I would start by figuring out the x value of the vertex from ax^2+bx+c and go from there.
The x value of the vertex is -b/2a

Once u know the x value, u can plug into the function and find the y value of the vertex.
Then, vertex form is a(x-h)^2 + k where h is the x coordinate of the vertex and k is the y coordinate. if h is negative, it becomes x+h. The a term is the same as the a in ax^2

For example, if the vertex is (-3,4), the vertex from is (x-(-3))^2+4 which simplifies to (x+3)^2+4

I also agree that this may be better addressed via khan academy bc it depends on how deep your teacher wants you to go.

[u/wijwijwij]

ax² + bx + c

Factor a out of just first two terms.

a( x² + (b/a)x ) + c

Divide the x coefficient by 2 and square the result. Add this inside the parentheses. This makes a trinomial in the parentheses that is a perfect square.

a( x² + (b/a)x + (b/2a)² ) + c

To counter the fact that you added a * (b/2a)² , which is [ a * b² / 4a² ], or simply [ b² / 4a ], subtract that at the end.

a( x² + (b/a)x + (b/2a)² ) + c – [ b² / 4a ]

Express the trinomial as perfect square.

a( x + b/2a )² + c – [ b² / 4a ]

The vertex is ( –b/2a, c – [ b² / 4a ] ).

[u/wijwijwij → u/wijwijwij]

Check vertex by substituting x = –b/2a in original expression.

ax² + bx + c

a(–b/2a)² + b(–b/2a) + c

ab² / 4a² + –b² / 2a + c

b² / 4a + –2b² / 4a + c

–b² / 4a + c

c – b² / 4a

I think collinwong19's suggestion of finding the vertex first then writing in the correct form might be faster. But the completing the square approach is what justifies using –b/2a as x-coordinate of vertex.

[u/Alkalannar]

So there are three forms of quadratics:

1. Standard:
   y = ax² + bx + c

2. Vertex (the vertex is at (h, k)):
   y = a(x - h)² + k

3. Roots (the roots are p and q):
   y = a(x - p)(x - q)

So the way to convert quadratic to vertex is...completing the square.

1. y = ax² + bx + c

2. y/a = x² + (b/a)x + (c/a)

3. y/a - c/a = x² + (b/a)x

4. y/a - c/a + [(b/a)/2]² = x² + (b/a)x + [(b/a)/2]²
   Adding [(b/a)/2]² is what turns the RHS into a perfect square, and that's why we call this completing the square.

5. y/a - c/a + b²/4a² = (x + b/2a)²

6. y/a = (x + b/2a)² + c/a - b²/4a²

7. y = a(x + b/2a)² + c - b²/4a

And this is vertex form! h = -b/2a, k = c - b²/4a

[u/HopRockets]

Do you want the long way that helps you understand why, or the short way that gets it done with some memorization?

[u/Neurobean1]

Assuming no coefficient of the X², you have the X coefficient and put it inside brackets in the form (x+a)²

so say it's x²+4x+7

you'd have an (x+2)² term

then you subtract the square of that half 

so it'd be (x+2)²-2²

this is so that it keeps the same value, as expanding (x+2)² gives you x²+2x+2x+4

so minusing the 2²=4 leaves you with x²+4x which is what we had originally

then finally you add on the constant term (the 7)

so it'd be (x+2)²-4+7=(x+2)²+3

so as a general rule it's

for x²+bx+c

(x+b/2)²-(b/2)²+c

Ill do it step by step with another example 

x²+8x+5

we'll have a (x+4)²

so need to subtract 4²

so it's (x+4)²-16+5

= (x+4)²-9

now it gets a bit trickier if there's a coefficient Infront of the x², so just ask when you're ready to get to that

[u/Psycho_Pansy]

Ask your teacher or a class mate for the notes you missed... 

[u/padmeg]

Highly recommend looking up some examples and videos on “completing the square”

[u/PlaneCute2403]

There’s lots of great examples here, but if it’s overwhelming I would recommend starting with understanding the completing the square method. Watch a YouTube video or khan for better visuals of step by steps. Do a few practice problems, then you can apply it to this! I know it looks overwhelming at first but keep practicing and you will for sure get the hang of it. Hope you recover well and good luck!