It is known that the image of quadratic function y = x & # 178; - 2x-3 intersects with X axis at points a and B, and the length of line AB is

It is known that the image of quadratic function y = x & # 178; - 2x-3 intersects with X axis at points a and B, and the length of line AB is

When y = 0, the solution is: X1 = 3, X2 = - 1, that is, the coordinate of point a is (3,0), and the coordinate of point B is (- 1,0)
So AB = 3 - (- 1) = 4
x\2+y\3=-3,7x=8y-4
Finding X and Y
x\2+y\3=-3,
Multiply both sides by 24
12x+8y=-72①
7x=8y-4
7x-8y=-4②
①+②
19x=-76
x=-4
Substituting (1)
y=-3
x=-4
y=-3
In the rectangular coordinate system, the length of the line AB cut by the vertex coordinates C (3,4) of the quadratic function image on the x-axis is 4
Analytic expression of 1 quadratic function
Two points P are on the parabola above the x-axis, and s △ PAB = 3S △ ABC, the coordinates of point P are obtained
3 find a little Q on the y-axis to minimize QA + QC
1. Vertex coordinates C (3,4), y = a (x-3) ^ 2 + 4, root is 3 + 2 / √ (- a), 3-2 √ (- a)
The length of line AB is 4,
That is, 4 = 4 / √ (- a), a = - 1
So y = - (x-3) ^ 2 + 4 = - x ^ 2 + 6x-5
2. Y = - (x-1) (X-5), abscissa 1 of point P (x, y)
1. If vertex C is (3,4), then the axis of symmetry is x = 3; if AB = 4, then point a is (1,0), and point B is (5,0)
Let the analytic formula of parabola be y = a (x-1) (X-5), and the image passes through point C (3,4), then:
4=a(3-1)*(3-5), a=-1.
So the analytic formula of quadratic function is: y = - (x-1) X-5) = - X & # 178; + 6x-5;
2. The expression of s △ PAB = 3S △ ABC is wrong, now it is changed to s △ PAB = (1 / 3) s △ ab
1. If vertex C is (3,4), then the axis of symmetry is x = 3; if AB = 4, then point a is (1,0), and point B is (5,0)
Let the analytic formula of parabola be y = a (x-1) (X-5), and the image passes through point C (3,4), then:
4=a(3-1)*(3-5), a=-1.
So the analytic formula of quadratic function is: y = - (x-1) X-5) = - X & # 178; + 6x-5;
2. The expression of s △ PAB = 3S △ ABC is wrong, now it is changed to s △ PAB = (1 / 3) s △ ABC
If cm is perpendicular to AB and PN is perpendicular to AB, then s ⊿ PAB / s ⊿ ABC = pN / cm
That is: (1 / 3) s ⊿ ABC] / s ⊿ ABC = pN / 4, 1 / 3 = pN / 4, PN = 4 / 3
4 / 3 = - X & # 178; + 6x-5, x = 3 + 2 √ 6 / 3 or x = 3-2 √ 6 / 3
So point P is (3 + 2 √ 6 / 3, 4 / 3) or (3-2 √ 6 / 3, 4 / 3)
3. Take the symmetric point a '(- 1,0) of point a (1,0) about y axis, then the intersection of a'c and Y axis is the required point Q
From the point a '(- 1,0) and point C (3,4), we can find that the line a'c is y = x + 1, so the point q is (0,1)
If 8y-7x = 0, then what is X: y?
8y-7x=08y=7x
x:y=8:7
There are not understand welcome to ask!
8y-7x=0
8y=7x
x:y=8:7
If you don't understand, you can ask
Hope to adopt
8y=7x
x:y=8:7
Let f (2) be the square root of the analytic function x (x) = 0
From F (x + 2) = f (2-x), we know that the axis of symmetry is x = 2, that is - B / 2A = 2 (1); X1 * X1 + x2 * x2 = (x1 + x2) * (x1 + x2) - 2x1 * x2 = b * B / (a * a) - 2C / a = 10 (2); 3 = C (3). Simultaneous (1), (2), (3), a = 1, B = - 4, C = 3, that is, f (x) = x * x-4x + 3
It is known that: 7x = 8x, then x: y = (), is x proportional to y?
It is known that: 7x = 8y, then x: y = (8:7), X and y are in (positive) proportion
7:8
Positive proportion
The quadratic function f (x + 3) = f (1-x), and the sum of squares of two real roots of F (x) = 0 is 10. If the image passes (0,3), find the analytic expression of F (x)
It should be solved by undetermined coefficient method
Let f (x) = ax ^ 2 + BX + C be substituted into f (x + 3) = f (1-x) to get a (x + 3) ^ 2 + B (x + 3) + C = a (1-x) ^ 2 + B (1-x) + C. according to Weida's theorem, X1 + x2 = - B / A, X1 * x2 = C / A, X1 ^ 2 + x2 ^ 2 = (B / a) ^ 2-2c / a = 10 (0,3), then f (0) = 0 + 0 + C = 3, the above equations are simultaneous, a, Ba = 1, B = - 4, C = 3
How to solve the quadratic equations of 2x-3y = 4,5x + 6y = 7?
2x-3y=4 (1)
5x+6y=7 (2)
(1)×2+(2)
4x+5x=8+7
9x=15
therefore
x=5/3
y=(2x-4)/3=-2/9
Using addition subtraction elimination method 2x-3y = 4 ①
5x+6y=7②
Solution 1 × 2 gives 4x-6y = 8, 3
② + 3 is 9x = 15
x=5/3
Substituting x = 5 / 3 into 1 gives 2 × 5 / 3-3y = 4
10/3-3y=4
-3y=2/3
y=-2/9
So x = 5 / 3, y = - 2 / 9
Let the quadratic function f [x] satisfy f [x + 2] = f [2-x], and the sum of squares of the two real roots of F [x] = 0 is 10, and the image of F [x] passes through the point [0,3], the solution of F [x] is obtained
How to get x = 2, what's the use
If the quadratic function f [x] satisfies f [a] = f [b],
Then, the axis of symmetry is x = (a + b) / 2
So the axis of symmetry is x = 2
Let f [x] = ax ^ 2 + BX + C
The image of F [x] passes through [0,3], so C = 3
Let two real roots be x1, x2
Then, X1 + x2 = - B / A, x1x2 = C / A
The axis of symmetry is x = 2
Then, X1 + x2 = 4,
The sum of squares of two real roots is 10
Then, (x1 + x2) ^ 2-2x1x2 = 10
That is, x1x2 = 3
arcsinx-x
b/a＝-4
c/a=3
C=3
The solution is a = 1, B = - 4
So, f [x] = x ^ 2-4x + 3
The answer is correct.
To solve the equations: 5x + 6y = 162x − 3Y = 1
5x + 6y = 16 & nbsp; & nbsp; ① 2x − 3Y = 1 & nbsp; & nbsp; ②, ① + ② × 2, 9x = 18, x = 2, 4-3y = 1, y = 1, so the solution of the equations is x = 2Y = 1