Question 356311
{{{f(x) = -x^2 + x -1}}}
X-intercepts are points where a graph intercepts/interesects the x axis. All points on the x axis have a y coordinate that is zero. (Think about it.) This tells us that X-intercepts have y coordinates that are zero. And since f(x) represents the y value, we want f(x) to be zero:
{{{0 = -x^2 + x -1}}}<br>
This is a quadratic equation. To solve a quadratic equation we first get one side of the equation to be zero. (We already have one side zero!). Then you either factor or use the Quadratic Formula. {{{-x^2 + x -1}}} does not factor easily we will use the Quadratic Formula:
{{{x = (-b +- sqrt(b^2 -4ac))/(2a)}}}
where the "a", "b" and "c" come from the general quadratic equation:
{{{ax^2 + bx + c = 0}}}
Looking at your equation we can see that
a = -1
b = 1
c = -1  (Be careful here. c = -1 because the general form has "+ c" at the end so we have to look at your equation as if there were all additions. So we have to look at " - 1" as if it were "+ (-1)". This is why c = -1.)
Subsituting these values for a, b and c into the formula we get:
{{{x = (-(1) +- sqrt((1)^2 - 4(-1)(-1)))/(2(-1))}}}
(Note how I used parentheses when I substituted. This is a very good habit to develop. It helps you avoid many kinds of errors that occur when you neglect to use parentheses.)
Simplifying we get:
{{{x = (-1 +- sqrt(1 - 4(-1)(-1)))/(-2)}}}
{{{x = (-1 +- sqrt(1 - 4))/(-2)}}}
{{{x = (-1 +- sqrt(-3))/(-2)}}}
At this point we should notice a problem. We have {{{sqrt(-3)}}} in our solution. But there are no Real numbers which can be squared to get -3. This means that there are no solutions to {{{0 = -x^2 + x -1}}}. And since there are no x values that make the y value 0, there are no points on the x axis (i.e. no x-intercepts). IOW, this parabola does not cross the x axis.<br>
Here's a graph so you can see this:
{{{graph(400, 400, -4, 4, -4, 4, -x^2+x-1)}}}