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what is the vertex and the x-intercepts of the function -2x^2+4x+6?

Guest Mar 21, 2015

Best Answer 

 #3
avatar+92623 
+10

$$\\y=-2x^2+4x+6\\
y=-2(x^2-2x-3)\\
$concave down because of the neg coeff of $ x^2\\
y=-2(x+1)(x-3)\\
$roots are x=-1 and x=3$\\
$axis of symmetry $x=\frac{-1+3}{2}=1\\
$Vertex y value$\\
y=-2*1^2+4*1+6)=-2+4+6=8\\
vertex \;\;(1,8)$$

Melody  Mar 22, 2015
 #1
avatar
+5

If you graph the equation (assuming that it is equal to zero), you will see that the x-intercepts are -1 and 3.

These can also be determined through the quadratic equation.

The vertex is (1,8), using -b/2a you can get the x-value of the vertex and then insert it into the equation to get the y-value.

Guest Mar 21, 2015
 #2
avatar+86859 
+5

Here's the graphical solution that confirms Anonymous' answers

https://www.desmos.com/calculator/fbm3sesyex

 

  

CPhill  Mar 22, 2015
 #3
avatar+92623 
+10
Best Answer

$$\\y=-2x^2+4x+6\\
y=-2(x^2-2x-3)\\
$concave down because of the neg coeff of $ x^2\\
y=-2(x+1)(x-3)\\
$roots are x=-1 and x=3$\\
$axis of symmetry $x=\frac{-1+3}{2}=1\\
$Vertex y value$\\
y=-2*1^2+4*1+6)=-2+4+6=8\\
vertex \;\;(1,8)$$

Melody  Mar 22, 2015

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