What is the vertexform y=-6x^2+3x+2 ?

The vertexform y=-6x^2+3x+2 is Maximum (1/4 , 19/8)

Popular >

Find the Vertex Form y=-6x^2+3x+2

vertex form

y = - 6 x^{2} + 3 x + 2

Maximum (\frac{1}{4}, \frac{19}{8})

Solution steps

y = - 6 x^{2} + 3 x + 2

Rewrite

y = - 6 x^{2} + 3 x + 2

in the vertex form

y = - 6 (x - \frac{1}{4})^{2} + \frac{19}{8}

The parabola parameters are:

a = - 6, h = \frac{1}{4}, k = \frac{19}{8}

Therefore the parabola vertex is

(h, k) = (\frac{1}{4}, \frac{19}{8})

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = - 6

Maximum (\frac{1}{4}, \frac{19}{8})

h (x) = x^{2} - 14 x + 6

8 x^{2} - 48 x + 65

x^{2} + y^{2} - 8 x - 6 y + 21 = 0

f (x) = x^{2} - 14 x + 6

f (x) = x^{2} - 6 x - 4

What is the vertexform y=-6x^2+3x+2 ?
The vertexform y=-6x^2+3x+2 is Maximum (1/4 , 19/8)