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標題:
中學的數學問題!!
發問:
y=2x''+4x ''=2次方 vertex??( , ) y-intercept??( , ) x-intercepts??( , )( , ) thz!!
最佳解答:
when y = 0, 2x^2 + 4x = 0 => x = 0 or x = -2 X-intercepts are 0,0 and -2,0 when x = 0 y = 0 Y-intercept is 0,0 d(y)/d(x) = 0 is the turning point => 4x + 4 = 0 => x = -1 when x = -1, y = -2 vertex is -1,-2
If you use the maths method, let y=2x^2+4x=0 vertex = -b/2a = -4/[2(2)] = -1 so vertex x-axis = -1 put x=-1 into y= 2x^2+4x y = 2(-1)^2+4(-1) = -2 so the curve vertex is (-1,-2) y-intercept is put x=0 into y= 2x^2+4x y = 2(0)^2+4(0) = 0 x-intercepts is put y=0 into y= 2x^2+4x 2x^2+4x = 0 x^2+2x = 0 x(x+2) = 0 x=0 or x=-2
中學的數學問題!!
發問:
y=2x''+4x ''=2次方 vertex??( , ) y-intercept??( , ) x-intercepts??( , )( , ) thz!!
最佳解答:
when y = 0, 2x^2 + 4x = 0 => x = 0 or x = -2 X-intercepts are 0,0 and -2,0 when x = 0 y = 0 Y-intercept is 0,0 d(y)/d(x) = 0 is the turning point => 4x + 4 = 0 => x = -1 when x = -1, y = -2 vertex is -1,-2
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其他解答:If you use the maths method, let y=2x^2+4x=0 vertex = -b/2a = -4/[2(2)] = -1 so vertex x-axis = -1 put x=-1 into y= 2x^2+4x y = 2(-1)^2+4(-1) = -2 so the curve vertex is (-1,-2) y-intercept is put x=0 into y= 2x^2+4x y = 2(0)^2+4(0) = 0 x-intercepts is put y=0 into y= 2x^2+4x 2x^2+4x = 0 x^2+2x = 0 x(x+2) = 0 x=0 or x=-2
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