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

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

Guest Mar 21, 2015

#3
+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
+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
+86859
+5

Here's the graphical solution that confirms Anonymous' answers

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

CPhill  Mar 22, 2015
#3
+92623
+10
$$\\y=-2x^2+4x+6\\ y=-2(x^2-2x-3)\\ concave down because of the neg coeff of  x^2\\ y=-2(x+1)(x-3)\\ roots are x=-1 and x=3\\ axis of symmetry x=\frac{-1+3}{2}=1\\ Vertex y value\\ y=-2*1^2+4*1+6)=-2+4+6=8\\ vertex \;\;(1,8)$$