Question

Plot the following?

`y=7-2x-x^2`

Answer 1

See below

Explanation:

`y=7-2x-x^2`

Since `y` is a quadraic function its graph will be a parabola

Let's reorder the terms into their usual form:

`->y = -x^2-2x+7`

Which is equivalent to `y = ax^2+bx+c`
Where: `a=-1, b=-2, c=+7`

a) We know that the parabola will have an axis of symmetry where `x=-b/(2a)`

i.e where `x= -(-2)/(2*-1) = -1`

b) Since, the coefficient of `x^2<0` `y` will have a maximum value on axis of symmetry.

Hence, `y_max = -(-1)^2-2(-1)+7`

`y_max = -1+2+7 =8`

`:. y` reaches a maximum at the point `(-1,8)`

c) `y` has zeros where `y=0`

i.e. where `-x^2-2x+7=0`

`:. x^2+2x-7=0`

Applying the quadratic formula

`x=(-2+-sqrt(2^2-4*1*(-7)))/(2*1)`

`= (-2+-sqrt(32))/2`

`= -1+-(4sqrt2)/2 = -1+-2sqrt2`

`x approx 1.8284 or -3.8284`

d) `y` will intercept the `y-`axis where `x=0`

i.e. where `y=-0-0+7 =7`

`:.` the `y-`intercept is the point `(0,7)`

Combining the results from a), b), c) and d) allows us to plot the graph of `y` as below.

graph{7-2x-x^2 [-28.86, 28.88, -14.44, 14.42]}

NB: In practice, to sketch the graph as shown we may need to compute and plot a few more points where `y<0`