Determine whether the graph of y = x^2 + 4x − 7 has a maximum or minimum point, then find the maximum or minimum value.

Minimum; (-11, -2)

Maximum; (-11, -2)

Maximum; (-2, -11)

Minimum; (-2, -11)

Question

Determine whether the graph of y = x^2 + 4x − 7 has a maximum or minimum point, then find the maximum or minimum value.

 Minimum; (-11, -2)

 Maximum; (-11, -2)

 Maximum; (-2, -11)

 Minimum; (-2, -11)

anonymous · Answer

same as last one

anonymous · Answer

its different though

hartnn · Answer

simply taking derivative and equating to 0 gives x=-2,and since double derivative=2=+ve,its a minimum

anonymous · Answer

leading coefficient is 1, so it opens up and has a minimum but no max

anonymous · Answer

min is at the vertex, first coordinate of the vertex is always 
$$-\frac{b}{2a}$$ in your case $a=1,b=4$ and so 
$$-\frac{b}{2a}=-\frac{4}{2}=-2$$

anonymous · Answer

ok so its minimum the first one

anonymous · Answer

oh no no be careful

anonymous · Answer

$-2$ is the first coordinate, not the second coordinate
you find the second coordinate by replacing $x$ by $-2$ to see what you get

anonymous · Answer

their is only two minimuns

anonymous · Answer

ohhhh
 gotcha

anonymous · Answer

can u continue to help me please

anonymous · Answer

if $x=-2$ then 
$$y=(-2)^2+4\times -2-7=-4-7=-11$$
\[

anonymous · Answer

so the vertex is $(-2,-11)$

anonymous · Answer

gotcha :)

anonymous · Answer

the question is worded incorrectly
i don't know where it came from, but it is not worded right
the MINIMUM value is a number, not a coordinate
the minimum value is $-11$

anonymous · Answer

http://openstudy.com/study#/updates/50256754e4b0d38989b743ff

anonymous · Answer

hmm idk