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german
3 years ago
8

What is the vertex form of y=2x^2-8x+1

Mathematics
1 answer:
bezimeni [28]3 years ago
6 0

Answer:

2(x-2)^2-7

Step-by-step explanation:

y=2x^2-8x+1

When comparing to standard form of a parabola: ax^2+bx+c

  • a=2
  • b=-8
  • c=1

Vertex form of a parabola is: a(x-h)^2+k, which is what we are trying to convert this quadratic equation into.

To do so, we can start by finding "h" in the original vertex form of a parabola. This can be found by using: \frac{-b}{2a}.

Substitute in -8 for b and 2 for a.

\frac{-(-8)}{2(2)}

Simplify this fraction.

\frac{8}{4} \rightarrow2

\boxed{h=2}

The "h" value is 2. Now we can find the "k" value by substituting in 2 for x into the given quadratic equation.

y=2(2)^2-8(2)+1

Simplify.

y=-7

\boxed{k=-7}

We have the values of h and k for the original vertex form, so now we can plug these into the original vertex form. We already know a from the beginning (it is 2).

a(x-h)^2+k\\ \\ 2(x-2)^2-7

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Step-by-step explanation:

We need to write equation of line perpendicular to y=0.25x-7 and passes through the point (-6,8)

Since the line y=0.25x-7 is in slope-inetercept form y=mx+b

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As, the new line is perpendicular to the given line so, slopes of lines that are perpendicular are: slope = -1/slope1

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Now the equation of the required line having slope m= -4 and b =30 will be:

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Keywords: Equation of line, slope-intercept form

Learn more about slope-intercept form at:

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According to the above, the mathematical expression would be:

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Answer:

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Step-by-step explanation:

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