Find the minimum y-value on the graph y=f(x)
f(x)= x^2+4x-5

Question

Find the minimum y-value on the graph y=f(x) 
f(x)= x^2+4x-5

anonymous · Answer

http://www.wolframalpha.com/input/?i=f%28x%29%3D+x%5E2%2B4x-5+

anonymous · Answer

if you want to know how to do it by hand...

you find the vertex. first step is to find the axis of symmetry, which is -b/2a.

once you find that, you know the x coordinate of the vertex (which is the lowest point in the graph, since it is a positive quadratic), so all you have to do is plug in your x to find the y coordinate.

anonymous · Answer

Well when I do it by how they are showing me what I am getting and what they are getting is two COMPLETELY different things.. extremely frustrating, I used the -6/2a

anonymous · Answer

well, you know that its not negative 6 over two a, its  negative b over two a. in this case,

a = 1
b = 4

anonymous · Answer

*-b/2a sorry

anonymous · Answer

then f(-2) followed by 
2^2+4*(-2)-5 correct therefore it gives you -1 ?

anonymous · Answer

not negative one. it should be negative 9. try your arithmetic again.

anonymous · Answer

and yet it says your answer is incorrect as well.

anonymous · Answer

maybe you typed the equation wrong? the absolute lowest y value for the equation x^2+4x-5 is -9. here's it's graph. https://www.google.com/search?q=x^2%2B4x-5&ie=utf-8&oe=utf-8&aq=t&rls=org.mozilla:en-US:official&client=firefox-a

anonymous · Answer

find $x=-\large{b\over2a}$ given that the equation is in $f(x)=ax^2+bx+c$ form.$$f(x)=x^2+4x-5$$$$x=-{4\over2(1)}$$$$x={-{4\over2}}$$$$x=-2$$now plug in this $x$ value into the equation and find $f(x)$:$$f(-2)=(-2)^2+4(-2)-5$$$$f(-2)=4+(-8)-5$$$$f(-2)=4-8-5$$$$f(-2)=-4-5$$$$f(-2)=-9$$ and thus the minimum value of $y$ is $-9$