How do you solve a/6=15/2?
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vertex = (3,- 5 )
given a quadratic in standard form : y = ax² + bx + c ( a ≠ 0 ), then
the x-coordinate of the vertex is
= -
y = x² - 6x + 4 is in standard form
with a = 1, b = - 6 and c = 4, hence
= -
= 3
substitute this value into the equation for y- coordinate
y = 3² - 6(3) + 4 = 9 - 18 + 4 = - 5
vertex = (3, - 5 ) → second table
x = 6
given f(x) = 310, we obtain the equation
10 + 50x = 310 ( subtract 10 from both sides )
50x = 300 ( divide both sides by 50 )
x = = 6