What is the vertex of #y=-(x + 2)^2 - 3x+9#?

Question

Genesis Allen · Accepted Answer

get the equation into the standard form of a quadratic

#y=ax^2+bx+c#

Expand the brackets

#y=-(x^2+4x+4)-3x+9#

Remove the brackets

#y=-x^2-4x-4-3x+9#

Collect like terms

#y=-x^2-7x+5#

Now use #(-b)/(2a)# to find the x coordinate of the vertex.

#(- -7)/(2xx -1)=7/(-2)#

Put this into the equation

#y=-(7/(-2))^2-7xx7/(-2)+5#

#y=-49/4+49/2+5#

#y=69/4#

The maximum is #(-7/2,69/4)#

Abigail Allen · Answer

undefined To find the vertex of the quadratic function $ y = -(x + 2)^2 - 3x + 9 $, first, rewrite the function in vertex form $ y = a(x - h)^2 + k $. Then, the vertex is at the point $(h, k)$. So, $ h $ is the x-coordinate of the vertex, and $ k $ is the y-coordinate of the vertex.

Abigail Allen · Answer

undefined To find the vertex of the quadratic function $ y = -(x + 2)^2 - 3x + 9 $, we need to rewrite the function in vertex form $ y = a(x - h)^2 + k $, where $ (h, k) $ represents the coordinates of the vertex.

By completing the square, we can rewrite the given function in vertex form:

$$ y = -(x + 2)^2 - 3x + 9 $$
$$ = -1(x^2 + 4x + 4) - 3x + 9 $$
$$ = -x^2 - 4x - 4 - 3x + 9 $$
$$ = -x^2 - 7x + 5 $$

Comparing this with the vertex form $ y = a(x - h)^2 + k $, we see that $ a = -1 $ and $ k = 5 $.

To find $ h $, we use the formula $ h = -\frac{b}{2a} $, where $ b = -7 $:

$$ h = -\frac{-7}{2(-1)} $$
$$ h = \frac{7}{2} $$

Therefore, the vertex of the function is $ \left(\frac{7}{2}, 5\right) $.