Interpreting that function as y=x2+2x+1, the graph of this function would be a parabola that opens upward. It would be equivalent to y=(x+1)2. Its vertex would be at (-1,0) and this vertex would be the parabola's only zero.
y=x2+3 x1=1 x2=2 y(x1) = 1*1+3 = 4 y(x2) = 2*2+3 = 7 x2/x1 = 2, While y2/y1 = 7/4 !=2, and thus the function is nonlinear.
1 and 2
It keeps the same shape and size, but the whole thing rises four units on the paper, as if by magic.
x-y = 1 => x = y+1 x2+y2 = 5 => (y+1)(x+1)+y2 = 5 2y2+2y-4 = 0 y = -2 or y = 1 So the points of intersection are: (-1, -2) and (2, 1)
Yes. Think of y as being a function of x. y = f(x) = x2 + 1
y = x2 + 2x + 1zeros are:0 = x2 + 2x + 10 = (x + 1)(x + 1)0 = (x + 1)2x = -1So that the graph of the function y = x + 2x + 1 touches the x-axis at x = -1.
Interpreting that function as y=x2+2x+1, the graph of this function would be a parabola that opens upward. It would be equivalent to y=(x+1)2. Its vertex would be at (-1,0) and this vertex would be the parabola's only zero.
y = x2 + 14x + 21 a = 1, b = 14 x = -b/2a = -14/2*1 = -7
No. x2+5x is a polynomial, an algebraic expression or a formula, but it is not a function. It could be used to help define a function. {(x,y) | y = x2 + 5x , x any real number} is a function
y = x2 + x = 0 x (X + 1) = 0 x = 0 is one solution x = -1 is the other
(xn+2-1)/(x2-1)ExplanationLet Y=1+x2+x4+...+xn. Now notice that:Y=1+x2+x4+...+xn=x2(1+x2+x4+...+xn-2)+1Y+xn+2=x2(1+x2+x4+...+xn-2+xn)+1Y+xn+2=x2*Y+1Y+xn+2-x2*Y=1Y-x2*Y=1-xn+2Y(1-x2)=1-xn+2Y=(1-xn+2)/(1-x2)=(xn+2-1)/(x2-1)
y=x2+2x+1 -b -2 2a= 2= -1 = axis of symmetry is negative one.
y = -x2 + 1 This function describes a parabola that opens downward. To find the top of it's range, you need to find it's focal point. You can do that very easily by taking the derivative of the equation and solving it for 0: y = -x2 + 1 ∴ y' = -2x let y' = 0: 0 = -2x ∴ x = 0 Now you can calculate the y value at that point: y = -02 + 1 ∴ y = 1 So that function describes an upside down parabola whose peak is at the point {0, 1}. It's range then is: {y | y ∈ ℜ, y ≤ 1}
y = x2 + 14x + 21; a = 1, b = 14, c = 21 x-coordinate = -b/2a = -14/2 = -7 if it is y = x2 + 14 + 21, the x-coordinate is zero, since b = 0.
There are several steps involved in how one can solve the derivative x plus y - 1 equals x2 plus y2. The final answer to this math problem is y'(x) = (1-2 x)/(2 y-1).
Proper form first. X2 + Y2 = 1 Y2 = 1 - X2 Y = (+/-) sqrt(1 - X2) -------------------------- zero out the X Y = (+/-) sqrt(1 - 02) Y = 1 ----------------the radius of this circle