how Do I find the X and Y intercepts of this equation? X(t)= -5x^2+105x+100?
回答 (4)
2019-02-27 11:39 pm
y = -5x² + 105x + 100
When x = 0:
y = -5(100)² + 105(0) + 100
y = 0
Hence, the y-intercept = 100
When y = 0:
0 = -5x² + 105x + 100
x² - 21x - 20 = 0
x = [21 ± √(21² + 4*20)]/2
x = (21 + √521)/2 or x = (21 - √521)/2
The x-intercepts = (21 + √521)/2 and x = (21 - √521)/2
When x = 0:
y = -5(100)² + 105(0) + 100
y = 0
Hence, the y-intercept = 100
When y = 0:
0 = -5x² + 105x + 100
x² - 21x - 20 = 0
x = [21 ± √(21² + 4*20)]/2
x = (21 + √521)/2 or x = (21 - √521)/2
The x-intercepts = (21 + √521)/2 and x = (21 - √521)/2
2019-02-27 11:19 pm
The way you wrote it, X(t) looks wrong. Do you mean y?
The y intercept is when x = 0. So just put in 0 instead of 'x', and figure out the value of y.
The x intercept is when y = 0. So solve 0 = -5x^2 + 105x + 100. You do remember the quadratic equation solution, don't you?
The y intercept is when x = 0. So just put in 0 instead of 'x', and figure out the value of y.
The x intercept is when y = 0. So solve 0 = -5x^2 + 105x + 100. You do remember the quadratic equation solution, don't you?
2019-02-27 11:49 pm
f(x) = -5x^2 + 105x + 100, x and y intercepts are ?
x = 0, f(x) = ?
f(0) = 0 + 0 + 100
f(0) = 100
y-intercept is 100.
f(x) = 0, x = ?
-5x^2 + 105x + 100 = 0
x^2 - 21x - 20 = 0
x = (21 ± √(521))/2
x-intercept are (21 - √(521))/2, (21 +√(521))/2.
x = 0, f(x) = ?
f(0) = 0 + 0 + 100
f(0) = 100
y-intercept is 100.
f(x) = 0, x = ?
-5x^2 + 105x + 100 = 0
x^2 - 21x - 20 = 0
x = (21 ± √(521))/2
x-intercept are (21 - √(521))/2, (21 +√(521))/2.
2019-02-28 12:30 am
y=-5x^2+105x+100
y=0=>
5x^2-105x-100=0=>
x=[105+/-sqr(-105)^2-4(5)(-100)]/[2(5)]=>
the x-intercept=21.9 or -.9 approximately.
x=0=>
the y-intercept=100
y=0=>
5x^2-105x-100=0=>
x=[105+/-sqr(-105)^2-4(5)(-100)]/[2(5)]=>
the x-intercept=21.9 or -.9 approximately.
x=0=>
the y-intercept=100
收錄日期: 2021-05-01 22:30:14
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https://hk.answers.yahoo.com/question/index?qid=20190227151007AAOOo5l