If two quadratic functions, images and shapes are the same, are their a equal?

If two quadratic functions, images and shapes are the same, are their a equal?

Quadratic function, f (x) = a x ^ 2 + BX + C
Can be reduced to f (x) = a (x + B / (2a)) ^ 2 + C-B ^ 2 / (4a ^ 2)
The form is a (x + m) ^ 2 + n
m. N is just the translation transformation in the horizontal and vertical coordinates of the image, while a determines the weight of the quadratic function, that is, the shape. The larger a is, the thinner the image is
So, a determines its shape, and B, C are just the translation of this quadratic function

How to prove that all quadratic functions have the same shape

All quadratic functions can become
The form of y = a (x-a) + B
We can see the translation of y = ax

Given the line segment Mn = 1, there is a point a on Mn, if an = 3 − 52

It is proved that: ∵ segment Mn = 1, there is a point a on Mn, an = 3 − 52, ∵ am = 1-3 − 52 = 5 − 12, ∵ AM2 = (5 − 12) 2 = 6 − 254 = 3 − 52, ∵ AM2 = an · Mn, ∵ point a is the golden section point of Mn

The shape problem of quadratic function image
Given that the shape of the image of the quadratic function is the same as y = - 2x ^ 2, ask whether the value of a of the quadratic function is - 2? Or | a | = | - 2 |, that is, a = 2 or - 2?

If the shape is the same as y = - 2x ^ 2, the absolute value of the coefficient of x ^ 2 is the same
So it's ± 2

Given the line segment Mn = 2, there is a point a on Mn. If an = 3-radical 5, is point a the golden section of line segment Mn?
Write process!

Ma = 2-3 + radical 5 = radical 5-1
MA / an = radical 5-1 / 3-radical 5 = (radical 5-1) (3 + radical 5) / 4 = (1 + radical 5) / 2
It's the golden section

When the image shape of quadratic function is reduced in proportion, the value of a will not change

It will change with the change of y value

Given the line segment Mn = 1, there is a point a on Mn, known an = (3-radical 5) / 2, please explain that point a is the golden section point of Mn

∵ Mn = 1, an = (3 - √ 5) / 2, am = 1 - (3 - √ 5) / 2 = (√ 5-1) / 2, am / Mn = (√ 5-1) / 2, an / am = (√ 5-1) / 2, am / Mn = an / am, that is, a M & # 178; = an × Mn

If the distance between M and N on the plane is 17cm, P is another point on the plane, and PM + PN = 25cm, then
1. Point P is on line Mn
2. Point P must be on line Mn
3. Point P is outside the line Mn
4. Point P must not be on line Mn

The point P must not be on the line Mn
Because if it is on the line Mn, then PM + PN = 17 ≠ 25cm

m. N is two points on the plane, Mn = 10cm, P is a point on the plane, if PM + PN = 20cm, then p is a point
A. It can only be outside the straight line Mn
B. Only on the line Mn
C. It can't be on the line Mn
D. Cannot be on line Mn

D

Find a point P, make PM = PN, and make the distance from point P to both sides of angle ABC equal

Point P is on the bisector of angle ABC
The distance from any point on the bisector to both sides of the corner is equal
It can be proved by congruent triangle. The angle bisector divides the angle into two equal angles, and then the distance between a point on the angle bisector and the two sides of the angle is drawn from the point to the edge. So it is an edge, angle bisector, and the vertical line from point to edge to form a right triangle. Two right triangles are congruent, so the two vertical lines from point to both sides are equal