For the function, f(x) = 5x²  10x + 6,
find the vertex and yintercept.
The function, f(x) = 5x²  10x + 6, is in standard form.
The yintercept is the constant term (the number without a variable), because at the yintercept, the value of x is equal to zero. The yintercept is 6. The xcoordinate of the vertex is b x =  2a In f(x) = 5x²  10x + 6 a = 5, b = 10, c = 6 (10) x =  {substituted 10 in for b and 5 in for a} 2(5) x = 1 {simplified} The xcoordinate of the vertex is 1. To find the ycoordinate of the vertex, substitute the xcoordinate, 1, back into the function. f(x) = 5x²  10x + 6 = 5(1)²  10(1) + 6 {substituted 1 in for x} = 5 + 10 + 6 {evaluated exponent and simplified} = 11 {simplified} The ycoordinate of the vertex is 11, which is also known as the maximum value of this particular function. The vertex is (1,11) Comments are closed.

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