Question

The equation of a 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

(i) the value of a and b,
(ii) the coordinates of the turning point of the curve

Checking answer ;; test tomorrow..........

Answer 1

if this can be written as
\[(x+4)^2+b\] then you have
\[x^2+8x+16+b=x^2+ax+3\]

Answer 2

this tells you that
\[8=a\] and
\[16+b=3\] so
\[b=-13\]

Answer 3

"turning point" (yikes) is the vertex. the graph is a parabola that opens up, so it "turns" at the vertex, when it goes from heading down (decreasing) to heading up (increasing)

Answer 4

we know that
\[b=-13\] and so we have
\[y=(x+4)^2-13\] making the vertex (-4,-13)
that is, since the first term
\[(x+4)^2\] is always greater than or equal to zero, it is smallest when it is 0.
it is 0 when x = -4, and if x = -4, then y = -13

Answer 5

thank you :'D