Question

The equation of the curve is given by `y=x^2+ax+3`, where `a` is a constant. Given that this equation can also be written as `y=(x+4)^2+b`, find (1) the value of `a` and of `b` (2) the coordinates of the turning point of the curve Someone can help?

The equation of the curve is given by `y=x^2+ax+3`, where `a` is a constant. Given that this equation can also be written as `y=(x+4)^2+b`, find

(1) the value of `a` and of `b`
(2) the coordinates of the turning point of the curve

Answer 1

The explanation is in the images.

Explanation:

Answer 2

`a=8,b=-13,(-4,-13)`

Explanation:

`x^2+ax+3to(1)`

`y=(x+4)^2+bto(2)`

`"expanding "(2)" using FOIL"`

`y=x^2+8x+16+b`

`color(blue)"comparing coefficients of like terms"`

`ax-=8xrArra=8`

`16+b-=3rArrb=3-16=-13`

`"the equation of a parabola in "color(blue)"vertex form"` is.

`color(red)(bar(ul(|color(white)(2/2)color(black)(y=a(x-h)^2+k)color(white)(2/2)|)))`

`"where "(h,k)" are the coordinates of the vertex and a"`
`"is a multiplier"`

`y=(x+4)^2-13color(blue)" is in vertex form"`

`rArrcolor(magenta)"vertex "=(-4,-13)larrcolor(blue)"turning point"`