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EastWind [94]
3 years ago
8

A teacher writes the function f(x)=3x^2-18x+27 on the board. How many times does the graph of the teacher's function intersect t

he x-axis
Mathematics
1 answer:
Oduvanchick [21]3 years ago
7 0

Answer:

Once. at (3,0)

Step-by-step explanation:

It is handy to have a graphing calculator for this kind of question.

Otherwise work it out.

All the terms are multiples of 3, so simplify.3(x^{2} -6x + 9)

then factor (x -3)(x-3)

Set each equal to 0 and solve for x.

x - 3 = 0    x = 3

OH ! they are identical! So there is only One value for x that will produce an output of 0.

So it intersects (touches) only once.

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\text{the three consecutive terms in an arithmetic sequence}\\a;\ a+d;\ a+2d\\\\\text{system of equals}\\\\ \left\{\begin{array}{ccc}a+a+d+a+2d=27\\a(a+d)(a+2d)=585\end{array}\right\\ \left\{\begin{array}{ccc}3a+3d=27&|:3\\a(a+d)(a+2d)=585\end{array}\right\\ \left\{\begin{array}{ccc}a+d=9&\to d=9-a\\a(a+d)(a+2d)=585\end{array}\right\\
\text{substitute to the second equation}\\\\a(a+9-a)(a+2(9-a))=585\\\\a(9)(a+18-2a)=585\\9a(18-a)=585\ \ \ |:9\\a(18-a)=65\\18a-a^2=65\ \ \ |\text{change the signs}\\a^2-18a=-65\\a^2-2a\cdot9=-65\ \ \ |+9^2\\\underbrace{a^2-2a\cdot9+9^2}_{(a-b)^2=a^2-2ab+b^2}=-65+9^2\\(a-9)^2=16\to a-9\pm\sqrt{16}\\a-9=-4\ \vee\ a-9=4\ \ \ |+9\\a=5\ \vee\ a=13\\\\d=9-5=4\ \vee\ d=9-13=-4
Answer:\ a=5;\ a+d=9; a+2d=13\ vee\ a=13;\ a+d=9;\ a+2d=5

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