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statuscvo [17]
3 years ago
7

Find the axis of symmetry and the vertex and graph the following: f(x)=(x+2)(x-4)

Mathematics
1 answer:
algol [13]3 years ago
8 0

something noteworthy is that the independent and squared variable in this case will be the "x", namely the graph of that quadratic is a vertical parabola.

\bf f(x) = (x+2)(x-4)\implies 0=(x+2)(x-4)\implies x = \begin{cases} -2\\ 4 \end{cases} \\\\\\ \boxed{-2}\rule[0.35em]{7em}{0.25pt}0\rule[0.35em]{3em}{0.25pt}\stackrel{\downarrow }{1}\rule[0.35em]{10em}{0.25pt}\boxed{4}

so the parabola has solutions at x = -2 and x = 4, and its vertex will be half-way between those two guys, namely at x = 1.

since this is a vertical parabola, its axis of symmetry, the line that splits its into twin sides, will be a vertical line, and it'll be the x-coordinate of the vertex, since the vertex hasa a coordinate of x = 1, then the axis of symmetry is the vertical line of x = 1.

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Answer:

answer = 12.87 km/h

Step-by-step explanation:

Given

Ship A is sailing east at 25 km/h = \frac{dx}{dt}

ship B is sailing north at 20 km/h =\frac{dy}{dt}

here x and y are the  sailing at t = 4 : 00 pm for ship A and B respectively

so we get x = 4 ×25 =100 km/h

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let z is the distance between the ships, we need to find \frac{dz}{dt} at t = 4 hr

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so equation will be

z^2 = (130-x)^2 + y^2......................(i)\\put x = 100 and y = 80 \\\\we |  | get \\

z^2 = 30^2 + 80^2\\z =\sqrt{7300} km/h

derivative first equation w . r. to t we get

2z\frac{dz}{dt} =-2(130-x)\frac{dx}{dt}+2y\frac{dy}{dt}

\frac{dz}{dt} =\frac{1}{z}[(x -130)\frac{dx}{dt} +y\frac{dy}{dt}]

\frac{dz}{dt} = \frac{( -20\times25 + 80\times20)}{\sqrt{7300} }

     = \frac{1100}{85.44}\\  = 12.87km/h

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