Question

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

Answer 1

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`