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Radda [10]
2 years ago
9

At how many points does the graph of the function below intersect the x axis? y= 16x2 - 8x+1

Mathematics
2 answers:
Radda [10]2 years ago
5 0
Y= 16x2 - 8x+1 does not intersect the c axis. It only peaks at it. The parabola does not go over it.
BartSMP [9]2 years ago
5 0
The vertex of the parabola occurs at (1/4, 0) or (0.25, 0). Since the coefficient of x^2 is a positive number, then it the parabola opens upward. The vertex is the minimum point on the graph, which happens to occur along the x-axis at (1/4, 0) as the only x-intercept.

Therefore, the graph intersects the x-axis only once, at the vertex.

Please mark my answers as the Brainliest if you find this helpful :)
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Perhaps you know that

S_2 = \displaystyle\sum_{k=1}^n k^2 = \frac{n(n+1)(2n+1)}6

and

S_3 = \displaystyle\sum_{k=1}^n k^3 = \frac{n^2(n+1)^2}4

Then the problem is trivial, since

\displaystyle\sum_{k=1}^n k^2(k+1) = S_2 + S_3 \\\\ = \frac{2n(n+1)(2n+1)+3n^2(n+1)^2}{12} \\\\ = \frac{n(n+1)\big((2(2n+1)+3n(n+1)\big)}{12} \\\\ = \frac{n(n+1)\big(4n+2+3n^2+3n\big)}{12} \\\\ = \frac{n(n+1)(3n^2+7n+2)}{12} \\\\ = \frac{n(n+1)(3n+1)(n+2)}{12}

Then

12\bigg(1^2\cdot2+2^2\cdot3+3^2\cdot4+\cdots+n^2(n+1)\bigg) = n(n+1)(n+2)(3n+1)

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