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How do you find the turnng point for y=-5t^2+4x+12?
1 Answer
- βαяяϵdθLv 79 years agoFavorite Answer
the vertex is your turning point . .
express this quadratic in vertex form then determine the vertex . .that is
y=-5t^2+4x+12 <---- y = a(x-h)^2 + k, where vertex at (h, k)
y = (5x^2 + 4x) + 12
y = 5(x^2 + 4/5*x) + 12
y = 5 (x^2 + 4/5*x + 4/25) + 12 - 4/5
y = 5 (x + 2/5)^2 + (60 - 4)/5
y = 5 (x + 2/5)^2 + 56/5
hence, vertex or turning point is
V (-2/5, 56/5) or in decimal form we have . ..
V (-0.4, 11.2)
