On the derivation of double radical of quadratic function Using the original method, find its derivation steps

On the derivation of double radical of quadratic function Using the original method, find its derivation steps

This is a (x \\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\] / 4A & # 178;] = a

Derivation of double radical of quadratic function
I want to know how to deduce the double radical of quadratic function
What is the basis
When we know the axis of symmetry, how do we know X1 + x2
If you know that the axis of symmetry is x = 4, find X1 + X1 =?
Why? 、
Is there a reason? 、

X1+X2=4*2=8
The basis is: (x1 + x2) / 2 = axis of symmetry

How to turn the vertex form of quadratic function into general form
emergency

y=a(x－h)²＋k
=a(x²－2hx＋h²)＋k
=Ax & # 178; - 2hax + ah & # 178; + K

What happens if the quadratic function y = (X-H) * 2 + K becomes y = (x + H) * 2 + k?
How will the vertex change after (h, K) becomes y = (x + H) * 2 + k? I've been thinking about it
There's nothing to think about. I'm just thinking about why it's subtraction, not addition. Why, if it's addition, it can be said that the peak is - H. Because generally speaking, the formula is good

The fixed point becomes (- H, K)
There's nothing to think about
2. It depends on the individual. You can calculate according to which method you think is easy to understand, but the current topics are all based on
Y = (X-H) ^ 2 + K, according to your method will produce a lot of errors
Is the vertex formula changed from the general formula, or is it calculated according to the original good formula

Is the vertex formula of quadratic function y = a (X-H) ^ 2 + K or y = a (X-H) 2 + k

Of course
y=a(x-h)^2+k
Vertex coordinates are (h, K)

What is the vertex of a quadratic function?

Because a quadratic function is a parabola, there are maxima and minima, which are vertices

How to change the general form of quadratic function into vertex form

Y = ax & # 178; + BX + C, which is changed into vertex form: y = a (x + B / 2a) &# 178; + (4ac-b & # 178;) / 4A
The formula process is as follows: y = ax & # 178; + BX + C
=a(x²+bx/a)+c
=a(x²+bx/a+b²/4a²-b²/4a²)+c
=a(x+b/2a)²-b²/4a+c
=a(x+b/2a)²+(4ac-b²)/4a
Hope to help you, if you do not understand, please hi me, I wish learning progress!

How can a quadratic function transform a general form into a vertex form,

y=ax²+bx+c=a[x²+ (b/a)x]+c=a[x²+2(b/2a)x]+c=a[x²+2(b/2a)x+(b/2a)²-(b/2a)²]+c=a[(x+b/2a)²-(b/2a)²]+c=a(x+b/2a)²-a(b/2a)²+c=a(x+b/2a)²-b²/4a+c...

The image of a quadratic function passes through (0,0), (- 1, - 1), (1,9) to find the analytic expression of the quadratic function
Let the analytic expression of quadratic function y = ax & # 178; + BX + C
Substituting (0,0), (- 1, - 1) and (1,9) to get:
c = 0
a - b + c = -1
a + b + c = 9
I don't understand this step at first
a = 4
b = 5
c = 0
So the analytic expression of quadratic function y = 4x & # 178; + 5x
The foundation is very poor. Can you introduce the elimination method in detail? Thank you for your help

From C = 0A - B + C = - 1 (1) a + B + C = 9 (2), substituting C = 0 into (1) and (2), A-B = - 1 (3) a + B = 9 (4) (3) formula + (4) formula 2A = - 1 + 9, so a = 4, substituting a = 4 into (4) formula 4 + B = 9, so B = 5, so a = 4B = 5C = 0, substituting a = 4, B = 5, C = 0 into y = ax & #

Given that the image of a quadratic function passes through (- 1, - 9), (1, - 3) and (3, - 5), the analytic expression of the quadratic function is obtained

Let the analytic expression of the quadratic function be y = AX2 + BX + C (a ≠ 0), substituting (- 1, - 9), (1, - 3) and (3, - 5), we get − 9 = a − B + C − 3 = a + B + C − 5 = 9A + 3B + C, and the solution is a = − 1b = 3C = − 5. So the analytic expression of the quadratic function is y = - x2 + 63x-5