SOLUTION: y=x^2/16+21x/16+11/8 the equation is x^2 devided by16 + 21x devided by 16 + 11 devided by 8 s a parabola. (i) Is the parabola u-shaped or n-

Algebra ->  Finance -> SOLUTION: y=x^2/16+21x/16+11/8 the equation is x^2 devided by16 + 21x devided by 16 + 11 devided by 8 s a parabola. (i) Is the parabola u-shaped or n-      Log On


   

Question 955425: y=x^2/16+21x/16+11/8 the equation is x^2 devided by16 + 21x devided by 16
+ 11 devided by 8
s a parabola.
(i) Is the parabola u-shaped or n-shaped? How can you tell this from
the equation? [1]
(ii) Use algebra to find the x-intercepts. [4]
(iii) Explain why the y-intercept is 11/8
(iv) Find the equation of the axis of symmetry, explaining your
method. Use this information to find the coordinates of the
vertex, rounding your answers where necessary to one decima i think the parabol is u shaped due to their being a negative number in the equation. i think the 2nd answer is y=x^2/16+21x/16+11/8 16y=x^2+21x+22
16y-x^2-21x-22=0 ax^2+bx+c=0 a=-1, b=-21, c=-22+16y
x=21 sqrt64y+353/-2 21-sqrt 64y+353/-2 simplyfy
x=21+sqrt 64y+353/2 - 21-sqrt 64y+353/2''.

Answer by MathLover1(20849) About Me  (Show Source):
You can put this solution on YOUR website!
y=x%5E2%2F16%2B21x%2F16%2B11%2F8+

(i) Is the parabola u-shaped or n-shaped? How can you tell this from
the equation?
the parabola is u-shaped, a=1%2F16 and a%3E0

(ii) Use algebra to find the x-intercepts.
y=x%5E2%2F16%2B21x%2F16%2B11%2F8+.... x-intercepts occur where y+=+0, so we have
0=x%5E2%2F16%2B21x%2F16%2B11%2F8+....both sides multiply by 16
0%2A16=16x%5E2%2F16%2B%2816%2A21x%29%2F16%2B%2816%2A11%29%2F8+
0=x%5E2%2B21x%2B2%2A11+
0=x%5E2%2B21x%2B22+....use quadratic formula
x+=+%28-21+%2B-+sqrt%28+21%5E2-4%2A1%2A22+%29%29%2F%282%2A1%29+
x+=+%28-21+%2B-+sqrt%28+441-88+%29%29%2F2+
x+=+%28-21+%2B-+sqrt%28+353+%29%29%2F2+.........exact solutions

x+=+%28-21+%2B-+18.79%29%2F2+
approximate solutions:
x+=+%28-21+%2B+18.79%29%2F2+
x+=-1.11
or
x+=+%28-21+-+18.79%29%2F2+
x+=-19.9



(iii) Explain why the y-intercept is 11%2F8
y-intercept occurs where x+=+0, yielding (0,11%2F8)
y=0%5E2%2F16%2B21%2A0%2F16%2B11%2F8+

(iv) Find the equation of the axis of symmetry, explaining your
method. Use this information to find the coordinates of the
vertex, rounding your answers where necessary to one decimal

The axis of symmetry is x+=+-b%2F2a. Since a=1%2F16 and b=21%2F16
x+=+-%2821%2F16%29%2F%282%281%2F16%29%29
x+=+-%2821%2Fcross%2816%29%29%2F%282%2Fcross%2816%29%29%29
x+=+-%2821%2F2%29
x+=+-10.5
That is also x-coordinate of the vertex, plug it in y=x%5E2%2F16%2B21x%2F16%2B11%2F8+ and find y-coordinate
y=%28-10.5%29%5E2%2F16%2B21%2A%28-10.5%29%2F16%2B11%2F8+
y=-5.5
the vertex is at: (-10.5,-5.5)