Answer to Question #186509 in Algebra for Petra

Question

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.

Answer 1

Solution.

$$2x^2-9x+10=\newline= 2(x^2-2\cdot\frac{9}{4}x+\frac{81}{16})-\frac{81}{8}+10=\newline =2(x-\frac{9}{4})^2-\frac{1}{8}.$$

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 $2x^2-9x+10=0$

$D=81-4•2•10=1,$

$x_1=\frac{9-1}{4}=2,\newline x_2=\frac{9+1}{4}=2.5.$

So, $(2,0),(2.5,0).$