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6 votes
Solve this equation
y=2x^(2)-4x+5

User Josejulio
by
7.9k points

2 Answers

10 votes

Answer:

Explanation:

Not sure what you mean by “solve,” since there are an infinite number of points on the curve.

y = 2x^2-4x+5 = 2(x^2 -2x) + 5 = 2(x^2-2x+1^2) -2(1^2) + 5 = 2(x-1)^2 + 3

This is an up-opening parabola with vertex (1,3).

User Nixmind
by
8.0k points
8 votes

Answer:

this is a function

Explanation:

ok so this we will solve with Order of Operations in mind, so first lets take a close look at the problem

y = 2x^{2} - 4x + 5

so since we have different variables on both sides of this equation, the answer isn't completely noticeable

since the right side of the equation is just a variable, lets try to substitute the x variable on the left side with a number, we will go with 3 for this time, so..

y = 2 * 3^{2} - 4 * 3 + 5

now solve for the number with the exponent

(3^{2}) = 9

y = 2 * 9 - 4 * 3 + 5

then following order of operations we will multiply...

2 * 9 = 18

4 * 3 = 12

y = 18 - 12 + 5

then add...

12 + 5 = 17

y = 18 - 17

then subtract...

18 - 17 = 1

y = 1

and VWOLA, when X = 3, Y = 1, but if we changed the value to say... 5, then we would do the problem all over again BUT all x values are = to 5, lets try it

x = 5

y = 2 * 5^{2} - 4 * 5 + 5

(5^{2}) = 25

y = 2 * 25 - 4 * 5 + 5

2 * 25 = 50

4 * 5 = 20

y = 50 - 20 + 5

20 + 5 = 25

y = 50 - 25

50 - 25 = 25

y = 25

and this is proof that this is a function, though it is not a LINIER function because the exponent causes the plotted line to curve on a graph

Solve this equation y=2x^(2)-4x+5-example-1
User Pise
by
7.4k points

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