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What is the vertex of the quadratic y -2x2 plus 4x-1?
It is (1, 1).
What is the range of y equals 2x2 plus 4x-3?
y = 2x2 + 4x - 3 This equation describes a parabola. Because it's first term is positive, we know that it goes infinitely upward, and has a minimum that occurs at it's vertex. You can find it's vertex by taking it's derivative and solving for zero: y' = 4x + 4 0 = 4x + 4 0 = x + 1 x = -1 y = (-1)2 + 4(-1) - 3 y = 1 - 4 - 3 y = -6 So the vertex is at (-1, -6), which means that y ≥ -6
What is the vertex of -2x2 plus 2x plus 3?
Since this question is in the calculus section, I'm assuming you know how to take the derivative. We know that y = -2x2 + 2x + 3 is a parabola, so it has one vertex, which is a minimum. We can use the first derivative test to find this extreme point.First, take the derivative:y' = -4x + 2Next, set y' equal to zero:0 = -4x + 2Then solve for x:4x = 2x = 2This is the x-coordinate of the vertex. To find the y-coordinate, plug x = 2 back into the original equation:y = -2x2 + 2x + 3y = -8 + 4 + 3y = -1So the vertex is at (2, -1).
What is the y-intercept of the graph of y -2x2-4x-3?
-3
What is the x-coordinate of the vertex of y equals 4x - 12 - 3?
y=4x-12-3 is the equation of a straight line. It does not have a vertex. Did you mean y=x squared - 12x - 3 ?
What is the vertex of y-2x2-4x-5?
2x2= 4
What is the vertex of the parabola y-2x2-4x 4?
Assuming the missing symbol there is an equals sign, then we have: y - 2x2 - 4x = 4 We can find it's vertex very easily by solving for y, and finding where it's derivative equals zero: y = 2x2 + 4x + 4 y' = 4x + 4 0 = 4x + 4 x = -1 So the vertex occurs Where x = -1. Now we can plug that back into the original equation to find y: y = 2x2 + 4x + 4 y = 2 - 4 + 4 y = 2 So the vertex is at the point (-1, 2)
What is the vertex of the quadratic y -2x2 plus 4x-1?
It is (1, 1).
What is the range of y equals 2x2 plus 4x-3?
y = 2x2 + 4x - 3 This equation describes a parabola. Because it's first term is positive, we know that it goes infinitely upward, and has a minimum that occurs at it's vertex. You can find it's vertex by taking it's derivative and solving for zero: y' = 4x + 4 0 = 4x + 4 0 = x + 1 x = -1 y = (-1)2 + 4(-1) - 3 y = 1 - 4 - 3 y = -6 So the vertex is at (-1, -6), which means that y ≥ -6
Which is the vertex for y equals 2x - 7?
-5
What is the vertex of the graph y-2x2 8x-3?
y=-2x^2+8x+3
What is the vertex of -2x2 plus 2x plus 3?
Since this question is in the calculus section, I'm assuming you know how to take the derivative. We know that y = -2x2 + 2x + 3 is a parabola, so it has one vertex, which is a minimum. We can use the first derivative test to find this extreme point.First, take the derivative:y' = -4x + 2Next, set y' equal to zero:0 = -4x + 2Then solve for x:4x = 2x = 2This is the x-coordinate of the vertex. To find the y-coordinate, plug x = 2 back into the original equation:y = -2x2 + 2x + 3y = -8 + 4 + 3y = -1So the vertex is at (2, -1).
What are the coordinates of the vertex of y x2 - 4x - 5?
Y = X2 - 4X - 5set to zeroX2 - 4X - 5 = 0X2 - 4X = 5halve the linear term ( - 4 ) then square it and add that result to both sidesX2 - 4X + 4 = 5 + 4factor on the left and gather terms together on the right(X - 2)2 = 9(X - 2)2 - 9 = 0==============vertex form(2, - 9)======vertex
What is the y-intercept of the graph of y -2x2-4x-3?
-3
What is the x-coordinate of the vertex of y equals 4x - 12 - 3?
y=4x-12-3 is the equation of a straight line. It does not have a vertex. Did you mean y=x squared - 12x - 3 ?
What is the vertex of 2x2 plus 4x?
We know that the x-coordinate of the vertex is x = -b/2a. We identify a, b, and c in f(x) = ax^2 + bx + c. y = 2x^2 + 4x or y = 2x^2 + 4x + 0 a = 2, b = 4, c = 0 Substitute 2 for a and 4 for b into the x-coordinate equation: x = -b/2a = -4/2(2) = -4/4 = -1 So the x-coordinate of the vertex is -1. To find the y-coordinate, substitute -1 for x in the equation of the function, y = 2x^2 + 4x. y = 2x^2 + 4x = 2(-1)^2 + 4(-1) = 2 - 4 = -2. Thus, the vertex is (-1, -2).
How do you write the function y equals 2 parenthese X-1 parenthese squared plus 3 in standard notation?
y = 2(x - 1)2 + 3 y = 2(x - 1)(x - 1) + 3 y = 2(x2 - 2x + 1) + 3 y = 2x2 - 4x + 2 + 3 y = 2x2 - 4x + 5
