What is the a value of this trinomial 4x^2+10x+24x 2 +10x+2?

Answer 1

y= 28x^2+20x+2, after combining like terms

in ax^2 +bx +c standard form, a=28, b=20, c=2

y'= 56x+20=0

x =-20/56 = -15/28= x coordinate of the vertex or minimum point

y= 28(15/28)^2 +20(-15/28)+2 =-19/28 = the y coordinate of the vertex (-15/28, -19/28)

y= 28(x+15/28)^2 +225/28-300/28+56/28

y= 28(x+15/28)^2 -19/28 in vertex form for an upward opening parabola

set y=0 and solve for x to find the zeros or x intercepts

y intercept =2

-15/28 +/- (1/28)sqr(400-8(28))= the zeros