What is the area of the triangle formed by the image of the linear function y = - 2X-4 and the two coordinate axes?

What is the area of the triangle formed by the image of the linear function y = - 2X-4 and the two coordinate axes?

y=0,x=-2
x=0,y=-4
So area = | - 2 | ×| - 4 | - 2 = 4
X = 0; y = - 4; y = 0; X = - 2 intersection X-axis at (- 2,0) intersection Y-axis at (0, - 4) = > area = 2 * 4 / 2 = 4
Four
Let the triangle area be s
Let y = - 2
Let x = 0, y = - 4
So the triangle area s = 2 | × | - 4 | △ 2 = 4
The first-order function y = - 2X-4 intersects the X axis at point a (- 2,0),
The intersection Y-axis is B (0, - 4),
So: the area of the triangle formed by the image of y = - 2X-4 and the two coordinate axes:
S△ABO = 1/2XOAXOB
= 1/2 X 2 X 4
= 4
Did you solve all these problems soon
Four
The triangle is a right triangle. Let x = 0, y = - 4; y = 0, x = - 2; area s = 1 / 2 * (2 * 4) = 4
Four
Set a = {a + radical 2B | a, B ∈ Q}, Q represents the set of rational numbers. If x1, X2 ∈ a, prove X1 + x2 ∈ a, X1 * x2 ∈ a
Because X1 and X2 belong to a, there are rational numbers A1, B1, A2 and B2, such that X1 = a1 + √ 2 * B1, X2 = A2 + √ 2 * B2, then X1 + x2 = (a1 + A2) + √ 2 * (B1 + B2), since a1 + A2 and B1 + B2 are rational numbers, X1 + x2 ∈ a; by X1 * x2 = (a1 + √ 2B1) (A2 + √ 2B2) = (A1A2 + 2b1b2) +
x1,x2∈A，
Let X1 = a + B √ 2, X2 = C + D √ 2,
Where a, B, C, D ∈ Q, then
x1+x2=（a+c)+(b+d)√2，
a+c,b+d∈Q,
∴x1+x2∈A;
x1x2=(ac+2bd)+(ad+bc)√2，
ac+2bd，ad+bc∈Q,
∴x1x2∈A.
The area of the triangle formed by the image and the two coordinate axes of the linear function y = - x + 2 is___ .
∵ let y = 0, then x = 2; let x = 0, then y = 2, ∵ for the image of the first-order function y = - x + 2, we can find the intersection point (2, 0) between the image and the X axis, and the intersection point with the Y axis is (0, 2) ∵ s = 12 × 2 × 2 = 2, so the answer is: 2
Set a = {x = a + B radical 2, a, B ∈ Z}, let x1 ∈ a, X2 ∈ a, prove: x1x2 ∈ a
The area of the triangle formed by the image of the first-order function passing through the point and surrounded by the two coordinate axes is 9 / 4, and the analytic expression of the first-order function is obtained
Let the expression of a linear function be y = KX + B, let x = 0, then: y = B; let y = 0, then: x = - B / K, that is: the intercept of a straight line on the X and Y axes are respectively: B and - B / K, and the triangle area enclosed by it and the coordinate axis is: 1 / 2 * b * | - B / K | = | B ^ 2 / 2K | = 9 / 4, then: k = 2B ^ 2 / 9 or K = - 2b ^ 2 / 9
There are many kinds of answers, and you can draw two kinds of pictures. I advise you, you are too lazy, the result is not so good, if you are willing to spend your time trying, the result will be very good, do you want to ask the idea or the process answer? The two decisions have different results
The two right angles of a triangle are related to the intersection of the first-order function and the X and Y axes. You don't say which point you go through, which can determine the positive or negative slope of a function. It is better to let the expression of the first-order function be y = KX + B.
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Set M = {x x = m + n radical 2 m ∈ Z n ∈ Z} X1 x2 belongs to m, then does X1 / X2 (x 2 ≠ 0) belong to m
not always
For example, X1 = 1 + √ 2
x2=2-√2
x1/x2=2+3√2/2
3 / 2 is not an integer
If the area of the intersection of △ A and △ B + y is known to be the Hohhot function (- 32m, y) and the Hohhot function (- 32m, y)
A. 2B. 3C. 4D. 6
The images of y = 32x + m and y = - 12x + n all pass through point a (- 2, 0), so we can get the expressions of 0 = 32 × (- 2) + m, 0 = - 12 × (- 2) + N, M = 3, n = - 1, respectively, y = 32x + 3, y = - 12x-1, y = 32x + 3 and y = - 12x-1, and the intersection points of Y axis and Y axis are B (0, 3), C (0, - 1), s △ AB, respectively
Given that f (x) = π ^ x, x1x2 > 0, compare the size of radical (f (x1) f (x2)) and f (radical (x1x2))
If you don't understand the title,
(1)
Radical (f (x1) f (x2))
=Radical (π ^ x1 × π ^ x2)
=Radical [π ^ (x1 + x2)]
=π^[1/2(x1+x2)]
=π ^ [radical (x1x2)]
(2)
π ^ x is an increasing function, similar to 2 ^ X
(3)
When both X1 and X2 > 0, X1 + x2 ≥ 2 radical (x1x2)
1 / 2 (x1 + x2) ≥ radical (x1x2)
(4)
When both X1 and X2 are less than 0, X1 + X2 is less than 0, and 2 radical (x1x2) is more than 0
So when X1 and X2 are all less than 0, X1 + X2 is less than 2 radical (x1x2)
1 / 2 (x1 + x2) < radical (x1x2)
Conclusion
When x 1 and x 2 are all greater than 0,
Radical (f (x1) f (x2)) ≥ f (radical (x1x2))
When both X1 and X2 are less than 0,
Radical (f (x1) f (x2)) < f (radical (x1x2))
The area of the triangle formed by the image and two coordinates of the linear function y = 3x + B is 24, and the value of B is calculated
The area of the triangle formed by the image and two coordinates of the linear function y = 3x + B
S=b^2/|2k|
So there is: B ^ 2 / | 2K | = B ^ 2 / 2x3 = 24
That is, B ^ 2 = 144
B = 12 or B = - 12
The intersection coordinates of the line and the X axis are (- B / 3,0), and the intersection coordinates of the line and the Y axis are (0, b). According to the triangle area formula: 1 / 2 * (|b | * | - B / 3 |) = 24, we can get B = 12 or B = - 12.
The triangle formed by the image of the function y = 3x + B and the two coordinates is a right triangle,
Then the image and two coordinates of the linear function y = 3x + B intersect at points a (0, b), B (- B / 3,0) respectively
The absolute value of the ordinate of point a is the length of one of the right angle sides|
The absolute value of the abscissa of point B is the length of another right angle side, | - B / 3|
So area = 1 / 2x | B | x | - B / 3 | = b * B / (2 * 3) = 24, so B = 12 or B = - 12... Expand
The triangle formed by the image of the function y = 3x + B and the two coordinates is a right triangle,
Then the image and two coordinates of the linear function y = 3x + B intersect at points a (0, b), B (- B / 3,0) respectively
The absolute value of the ordinate of point a is the length of one of the right angle sides|
The absolute value of the abscissa of point B is the length of another right angle side, | - B / 3|
So area = 1 / 2x | B | x | - B / 3 | = b * B / (2 * 3) = 24, so B = 12 or B = - 12
Given that f (x) = π ^ x, x1x2 > 0, try to compare the size of F (x1). F (x2) under the root sign with that of x1x2 under the root sign
f(x1).f(x2)=π^(x1+x2)
X1x2 under f radical = x1x2 under π ^ radical
F (x1). Under f (x2) - f radical, x1x2 = π ^ (x1 + x2) - π ^ radical, x1x2
F (x) = π ^ x is an increasing function
Compare X1 + x2 with x1x2 under root sign
If X1 + X2 is greater than = 2 times the root sign, x1x2 is greater than x1x2 under the root sign
So f (x1). Under f (x2) - f radical, x1x2 = π ^ (x1 + x2) - π ^ radical, x1x2 is greater than 0
That is, f (x1). F (x2) is greater than x1x2 under f radical