# Answer in Algebra for Petra #186509

a. Use complete square method to solve for x and sketch (x)=2x^2-9x+10

b. What is the range of the parabola?

c. What is the domain of the parabola?

d. Sketch your graph and label your vertex, x intercept and y intercept.

e. Use the Quadratic formula to verify the x intercepts.

Solution.

“2x^2-9x+10=newline= 2(x^2-2cdotfrac{9}{4}x+frac{81}{16})-frac{81}{8}+10=newline =2(x-frac{9}{4})^2-frac{1}{8}.”

The graph of* *“y=2(x-frac{9}{4})^2-frac{1}{8}” is congruent to the basic parabola “y=x^2” , but is translated “frac{9}{4}” units to the right, “frac{1}{8}” units down and

compression 2 times to the y-axis.

Domain = R.

Range = “(-frac{1}{8}, infty).”

Vertex “(frac{9}{4},-frac{1}{8}).”

Y intercept “(0,10).”

X intercept “2x^2-9x+10=0”

“D=81-4u20222u202210=1,”

“x_1=frac{9-1}{4}=2,newlinenx_2=frac{9+1}{4}=2.5.”

So, “(2,0),(2.5,0).”