mathdad Full Member Joined Apr 24, 2015 Messages 941 May 7, 2019 #1 Find the vertex: -4x^2 + 16x - 7 Vertex = ( x, f(x)). x = -b/2a x = -16/2(-4) x = -16/-8 x = 2 f(x) = -4x^2 + 16x - 7 Let x = 2 f(2) = -4(2)^2 + 16(2) - 7 f(2) = -4(4) + 32 - 7 f(2) = -16 + 32 - 7 f(2) = 16 - 7 f(2) = 9 Vertex = (2, 9) Yes?
Find the vertex: -4x^2 + 16x - 7 Vertex = ( x, f(x)). x = -b/2a x = -16/2(-4) x = -16/-8 x = 2 f(x) = -4x^2 + 16x - 7 Let x = 2 f(2) = -4(2)^2 + 16(2) - 7 f(2) = -4(4) + 32 - 7 f(2) = -16 + 32 - 7 f(2) = 16 - 7 f(2) = 9 Vertex = (2, 9) Yes?
MarkFL Super Moderator Staff member Joined Nov 24, 2012 Messages 3,021 May 7, 2019 #2 [MATH]f(x)=-4x^2+16x-7=-4(x^2-4x)-7=-4(x^2-4x+4)-7+16=-4(x-2)^2+9[/MATH] Vertex is: [MATH](2,9)\quad\checkmark[/MATH]
[MATH]f(x)=-4x^2+16x-7=-4(x^2-4x)-7=-4(x^2-4x+4)-7+16=-4(x-2)^2+9[/MATH] Vertex is: [MATH](2,9)\quad\checkmark[/MATH]
