Asked by kalum
"A woodland jumping mouse hops along a parabolic path given by y=-0.2x(squared)+1.3x where x is the mouse's horizontal position(in feet) and y is the corresponding height(in feet). Can the mouse jump over a fence that is 3 feet high? Explain in quadratic function."
Answers
Answered by
Reiny
so you would want
3 = -.2x^2 + 1.3x
.2x^2 - 1.3x + 3 = 0
times 10
2x^2 - 13x + 30 = 0
x = (13 ±√-71)/4
which is not a real number,
so NO, the mouse cannot jump over that fence
or
the x of the vertex is 1.3/(2(.2)) = 1.3/.4 = 13/4
when x = 13/4
y = -.2(13/4) + 1.3(13/4) = appr 2.1
since the parabola opens downwards the max value of y is 2.1
but we needed y = 3
NO WAY!
3 = -.2x^2 + 1.3x
.2x^2 - 1.3x + 3 = 0
times 10
2x^2 - 13x + 30 = 0
x = (13 ±√-71)/4
which is not a real number,
so NO, the mouse cannot jump over that fence
or
the x of the vertex is 1.3/(2(.2)) = 1.3/.4 = 13/4
when x = 13/4
y = -.2(13/4) + 1.3(13/4) = appr 2.1
since the parabola opens downwards the max value of y is 2.1
but we needed y = 3
NO WAY!
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