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 looks like a parabola which looks like a U shape.
It consists of two disjointed line segments: x ≤ -3 and x ≥ 3.
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
Rise divided by run. (Y2 - Y1) / (X2 - X1) - with (X1, Y1) and (X2, Y2) being two points on the graph.
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.
The graph is a parabola facing (opening) upwards with the vertex at the origin.
y = (square root 1- x2) + (cube root x2)