Question

f(x) = 2x² − 5x + 3

1) What are the x-intercepts of the graph of f(x)? Show your work.

2) Is the vertex of the graph of f(x) going to be a maximum or minimum? What are the coordinates of the vertex? Justify your answers and show your work.

3) What are the steps you would use to graph f(x)? Justify that you can use the answers obtained in Part A and Part B to draw the graph.

Answer 1

Part-1

.

.

$y= 2x^2-5x+ 3$

to find the x-intercept take y=0

$2x^2-5x+ 3=0$

$2x^2-3x-2x+3=0$

$x(2x-3)-1(2x-3)=0$

$(2x-3)(x-1)=0$

$x=\frac{3}{2},\:x=1$

${\color{Red} \left(\frac{3}{2},\:0\right),\:\left(1,\:0\right)}$

${\color{Red} (1.5,\:0),\:\left(1,\:0\right)}$..............x-intercept

.

.

.

to find the y-intercept take x=0

$y=2\cdot \:0^2-5\cdot \:0+3$

$y=0-0+3$

$y=3$

.

${\color{Red} \left(0,\:3\right)}$..............y-intercept

.

.

Part-2

.

.

$y= 2x^2-5x+ 3$

$y=2\left(x^2-\frac{5x}{2}+\frac{3}{2}\right)$

to convert into vertex form add and subtract  $\frac{25}{16}$

$y=2\left(x^2-\frac{5x}{2}+\frac{3}{2}+\frac{25}{16}-\frac{25}{16}\right)$

$y=2\left(x^2-\frac{5x}{2}+\frac{25}{16}+\frac{3}{2}-\frac{25}{16}\right)$

here we can write $x^2-\frac{5x}{2}+\frac{25}{16}=\left(x-\frac{5}{4}\right)^2$

.

$y=2\left(\left(x-\frac{5}{4}\right)^2+\frac{3}{2}-\frac{25}{16}\right)$

$y=2\left(\left(x-\frac{5}{4}\right)^2-\frac{1}{16}\right)$

$y=2\left(x-\frac{5}{4}\right)^2-2\cdot \frac{1}{16}$

$y=2\left(x-\frac{5}{4}\right)^2-\frac{1}{8}$

compare with vertex form of the parabola  $y=a\left(x-h\right)^2+k$

so here vertex is ${\color{Red} \left(h,\:k\right)=\left(\frac{5}{4},\:-\frac{1}{8}\right)=(1.25, \:-0.125)}$

.

.

here coefficient of x² is 2, which is positive so vertex is minimum

.

.

Part-3

.

.

here vertex ${\color{Red} \left(h,\:k\right)=\left(\frac{5}{4},\:-\frac{1}{8}\right)=(1.25, \:-0.125)}$ is minimum

so parabola opens upward

so draw a parabola with the help of x-intercept and vertex which is minimum

0.5 (1,0) (1.5, 0) (1.25. -0.125)