How do I find the vertex, axis of symmetry, y-intercept, x-intercept, domain and range of #v(t) = t^2 + 11t - 4#?

1 Answer
iceman
Sep 29, 2015

Please see the solution below.

Explanation:

#f(x)=ax^2+bx+c=>#the general/standard form of the quadratic function, it is represented by the graph of the parabola where the coordinate of the vertex is:
#[-b/(2a), f(-b/(2a))]# , so in this case we have:
#v(t)=t^2+11t-4# , then:
#t=-11/2#
#v(-11/2)=(-11/2)^2+11(-11/2)-4#
#=121/4-121/2-4=-137/4#
#:.# the vertex is at #(-11/2, -137/4)#

The axis of symmetry is the same as the x-coordinate of the vertex, so in this case:
#t=-11/2=># axis of symmetry

To find the y-intercept set x to zero and solve for y, in this case:
#v=0+0-4=-4=>#the y-intercept is at #(0, -4)#

To find the x-intercepts set y to zero and solve for x:
#t^2+11t-4=0=># solution by completing the square:
#t^2+11t+(11/2)^2=4+121/4#
#(t+11/2)^2=137/4#
#t+11/2=+-sqrt137/2#
#t=-11/2+-sqrt137/2=># hence the x-intercepts are:
#(-11/2+-sqrt137/2 , 0)#

The domain of parabola is all the real numbers, but the range is limited by y-coordinates of the vertex, in this case since the parabola opens up the vertex is the minimum so the range is:
#v>=-137/4#

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