Question

Determine whether f(x)=-5x^2-10x+6 has a maximum or a minimum value. Find that value and explain how you know.

Answer 1

Answer:

(-1, 11) is a max value; parabola is upside down

Step-by-step explanation:

We can answer this question backwards, just from what we know about parabolas.  This is a negative x^2 parabola, so that means it opens upside down.  Because of this, that means that there is a max value.  

The vertex of a parabola reflects either the max or the min value.  In order to find the vertex, we put the equation into vertex form, which has the standard form:

$y=a(x-h)^2+k$

where h and k are the coordinates of the vertex.

To put a quadratic into vertex form, you need to complete the square.  That process is as follows. First, set the quadratic equal to 0.  Then make sure that the leading coefficient is a positive 1.  Ours is a -5 so we will have to factor it out.  Then, move the constant to the other side of the equals sign.  Finally, take half the linear term, square it, and add it to both sides.  We will get that far, and then pick up with the rest of the process as we come to it.

$-5x^2-10x+6=y$

Set it to equal zero:

$-5x^2-10x+6=0$

Now move the 6 to the other side:

$-5x^2-10x=-6$

Factor out the -5:

$-5(x^2+2x)=-6$

Take half the linear term, square it, and add it to both sides.  Our linear term is 2x.  Half of 2 is 1, and 1 squared is 1, so add it to both sides.  Keep it mind that we have the =5 out front of those parenthesis that will not be forgotten.  So we are not adding in a +1, we are adding in a (+1)(-5) which is -5:

$-5(x^2+2x+1)=-6-5$

In completing the square, we have created a perfect square binomial on the left.  Stating that binomial along with simplifying on the right gives us:

$-5(x+1)^2=-11$

Now, bring the -11 over to the other side and set it back to equal y and you're ready to state the vertex:

$-5(x+1)^2+11=y$

The vertex is at (-1, 11)