How do you use the graphing utility to graph #f(x)=-2x^2+10x# and identify the x intercepts and vertex?

1 Answer
Jan 14, 2017

Answer:

Vertex : (2.5, 12,5). x-intewrcept ( y = 0 ) : 5 , See graph and explanation

Explanation:

graph{-2x^2+10x [-40, 40, -20, 20]} y = a quadratic in x and x = a quadratic in y represent parabolas, with

axis parallel to y/x-axis, respectively.

In the standard form, the given equation is

#(x-2.5)^2=-1/2(y-12.5)#, showing

vertex at V(2,5, 12.5) and

axis parallel to y-axis #darr#.

The local graph below includes the axis and the tangent at the

vertex that meet at the vertex V(2.5, 12,5).

In this not-to-scale,I could see my y-scale causing some problems.

For contrast, I have included a uniform-scale graph.

graph{(y+2x^2-10x)((x-5)^2+(y-12.5)^2-.1)((x-5)^2+y^2-.1)((x-10.8)^2+y^2-.1)(y-12.5)(x-2.5)=0 [-10, 10, -100, 100]}