# what is a ? ( a,-5) y=x^2+6x must show work

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Considering the point (a,-5) a point that lies on the graph of the given function, hence its coordinates verify the equation of the function such that:

`-5 = a^2 + 6a `

Notice that you need to substitute -5 for y and a for x.

`a^2 + 6a + 5 = 0`

You need to use quadratic formula such that:

`a_(1,2) = (-6+-sqrt(36 - 20))/2`

`a_(1,2) = (-6+-sqrt16)/2 => a_1 = (-6+4)/2 => a_1 = -1`

`a_2 = (-6-4)/2 => a_2 = -5`

**Hence, evaluating the x coordinate a yields (-1,-5) and (-5,-5) for a = -1 and a = -5.**

Substitute x=4 and y=-5 in the given equation y=x^2+6x and solve for a by factoring

-5=a^2+6a

a^2+6a+5=0

(a+1)(a+5)=0

therefore, a=-1 or a=-5

if you draw the graph (paraboll)it will pass (-1,5) and (-5,5) i.e axis of symetry x=(-1-5)/2=-3