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Minimum or maximum value Equation in vertex form y=- (x+5)^2 -3Asked by hala aly
Minimum or
maximum value Equation in
vertex form 𝒚 = 𝟑𝒙^2+ 4
maximum value Equation in
vertex form 𝒚 = 𝟑𝒙^2+ 4
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The minimum or maximum value of an equation in vertex form y = a(x-h)^2 + k can be found by looking at the value of a and the sign of a.
If a is positive, the vertex of the parabola is a minimum point and the minimum value of the equation is y = k.
If a is negative, the vertex of the parabola is a maximum point and the maximum value of the equation is y = k.
Using the given equation in vertex form y = 3x^2 + 4, we can see that a is positive (3 is positive). Therefore, the vertex is a minimum point and the minimum value of the equation is y = 4.
So the minimum value of the equation y = 3x^2 + 4 is 4.
If a is positive, the vertex of the parabola is a minimum point and the minimum value of the equation is y = k.
If a is negative, the vertex of the parabola is a maximum point and the maximum value of the equation is y = k.
Using the given equation in vertex form y = 3x^2 + 4, we can see that a is positive (3 is positive). Therefore, the vertex is a minimum point and the minimum value of the equation is y = 4.
So the minimum value of the equation y = 3x^2 + 4 is 4.