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Mathematics 22 Online
OpenStudy (anonymous):

what is the vertex of the graph of y=2x^@+4x-5 a)(2,11) b) (1,) c) (0,-5) d) (-1,-7)

OpenStudy (anonymous):

@skinny23 @Eyad @amistre64 @hartnn some one help please

OpenStudy (zpupster):

the quadratic in this form-- ax^2+bx+c uses the formula -b/2a so from your equation : a=2 and b = 4 so, -(4)/2(2)= -4/4= -1 now plug back in -1 2(-1) ^2+ 4(-1) -5 2-4-5 = -7 your vertex is at -1,-7

OpenStudy (anonymous):

heres anopther one y=-2x^2+8x+9 a)(2,17) b)(1,15) c)(0,9) d) (-1,-1)

OpenStudy (anonymous):

@zpupster

OpenStudy (zpupster):

again just try plugging in a = -2 b= 8 use the formula for finding x value of vertex (or usually they use h) -b/2a -(8)/2(-2) = -8/-4= 2 I will let you plug that in the formula to get the y (or k value)

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