If f(x) = x2 + 25, then to plot f(x) on a graph would give you a parabolic curve extending infinitely upward with a minimum value of 25, and it's vertex at the point (0, 25).
It looks like a parabola which looks like a U shape.
Assuming you meant y=x2 & y=x2-4 They are both straight-line graphs, however - they produce different results. Using the values of 1,2,3,4 & 5 for x (as an example)... In the first equation, the value of y would be 1,4,8,16 & 25 In the second equation, y would be -3,0,4,12 & 21
the graph is moved down 6 units
x2+(y-x2/3)2=1
y=x2+4x+1
if y = x2 + 10x + 25 then y = (x + 5)2 This tells us that the graph would be a parabola, with it's vertex at (-5, 0), and a range of 0 to infinity.
No translation will invert a quadratic graph.
It consists of two disjointed line segments: x ≤ -3 and x ≥ 3.
It looks like a parabola which looks like a U shape.
Assuming you meant y=x2 & y=x2-4 They are both straight-line graphs, however - they produce different results. Using the values of 1,2,3,4 & 5 for x (as an example)... In the first equation, the value of y would be 1,4,8,16 & 25 In the second equation, y would be -3,0,4,12 & 21
the graph is moved down 6 units
x2+(y-x2/3)2=1
y=x2+4x+1
Select two points on the graph and suppose their coordinates are (x1, y1) and (x2, y2) then the gradient = (y1 - y2) / (x1 - x2) provided that x1 and x2 are different. If not, the gradient is not defined.
Rise divided by run. (Y2 - Y1) / (X2 - X1) - with (X1, Y1) and (X2, Y2) being two points on the graph.
The graph is a parabola facing (opening) upwards with the vertex at the origin.
y = (square root 1- x2) + (cube root x2)