The equation of a quadratic in standard form is: y=ax^2+bx+cy=ax2+bx+c
So, this question is asking us to find a, b , ca,b,c
y=(2x+14)(x+12) - (7x-7)^2y=(2x+14)(x+12)−(7x−7)2
It is probably simplier to break yy in its two parts first.
y = y_1 - y_2y=y1−y2
Where: y_1 = (2x+14)(x+12)y1=(2x+14)(x+12) and y_2= (7x-7)^2y2=(7x−7)2
Now, expand y_1y1
y_1 = 2x^2+24x+14x+168y1=2x2+24x+14x+168
= 2x^2+38x+168=2x2+38x+168
Now, expand y_2y2
y_2 = (7x-7)^2 = 7^2(x-1)^2y2=(7x−7)2=72(x−1)2
=49(x^2-2x+1)=49(x2−2x+1)
= 49x^2-98x+49=49x2−98x+49
We can now simply combine y_1 - y_2y1−y2 to form yy
Thus, y= 2x^2+38x+168 -(49x^2-98x+49)y=2x2+38x+168−(49x2−98x+49)
= 2x^2+38x+168 -49x^2+98x-49=2x2+38x+168−49x2+98x−49
Combine coefficients of like terms.
y = (2-49)x^2 + (38+98)x +(168-49)y=(2−49)x2+(38+98)x+(168−49)
y= -47x^2 + 136x +119y=−47x2+136x+119 (Is our quadratic in standard form)
a=-47, b=+136, c=+119a=−47,b=+136,c=+119
