What is the vertex form of $y=-x^2-3x+5$ ?

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Answer 1 · 2015-12-25T11:24:21

There are many ways of finding the vertex form of this type quadratic functions. An easy method is given below.

If we have $y =ax^2+bx+c$ and to write it in vertex form we do the following steps.

If the vertex is $(h,k)$ then $h=(-b/(2a))$ and $k=a(h)^2+b(h)+c$

The vertex form is y=a(x-h)^2 + k.

Now let us use the same with our question.

$y=-x^2-3x+5$

Comparing it with $y=ax^2+bx+c$ we get $a=-1$ , $b=-3$ , $c=5$

$h=-b/(2a)$
$h=-(-3)/(2(-1))$
$h=-3/2$

$k=-(-3/2)^2-3(-3/2)+5$
$k=-9/4 +9/2 + 5$
$k=+9/4 + 5$
$k= 9/4 + 20/4$
$k=29/4$

$y=-(x-(-3/2))^2 + 29/4$

$y=-(x+3/2)^ 2+ 29/4$ is the vertex form

What is the vertex form of y=-x^2-3x+5y=-x^2-3x+5?