Find the x-intercepts. y = 2x^2 - 9x + 1?

Question

Need some more Algebra Help if anyone would mind helping a girl out.

Thanks!

Find the x-intercepts. y = 2x^2 - 9x + 1

Answer 1

The x-intercepts are the points where the function crosses the x-axis. On the x-axis, all y-coordinates are zero. (Positive and they would be above the x-axis, negative and they would be below, but zero they are on the line).

So you just need to find out what values of x will make the y value = 0. So simply set y = 0.

y = 2x² - 9x + 1
0 = 2x² - 9x + 1

At this point you need to factor this, but it won't come out with integer coefficients... so let's use the quadratic formula:

In general, when you have:
0 = ax² + bx + c

Then the formula for x is:
..... -b ± sqrt( b² - 4ac )
x = --------------------------
................ 2a

In your case: a = 2, b = -9, c = 1

Plug these in:
..... -(-9) ± sqrt( (-9)² - 4(2)(1) )
x = ------------------------------------
................... 2(2)

Simplify:
..... 9 ± sqrt( 81 - 8 )
x = ----------------------
.................. 4

..... 9 ± sqrt( 73 )
x = ------------------
.................. 4

So that's about as far as you can get. Your two x-values are:
[9 + sqrt(73) ] / 4
or
[9 - sqrt(73) ] / 4

Therefore the two x-intercepts are the following points:
( [9 + sqrt(73) ] / 4, 0 )
( [9 - sqrt(73) ] / 4, 0 )