For what values of m does the graph of y = mx2 – 5x – 2 have no x-intercepts?

"no x-intercept(s)" => graph never even touches the x-axis.

y = mx^2 – 5x – 2  is the equation of a parabola that opens up.

The quadratic formula would be useful here.  Note that there are 3 possible outcomes when this formula is applied:  the discriminant b^2 - 4ac could be zero, positive or negative.  If the discrim. is negative, then the roots (solutions) are complex; the graph does not touch the x-axis.

So, let's find the discrim. of  y = mx2 – 5x – 2.  here a=m, b= -5 and c= -2.  The discrim. is b^2 - 4(a)(c), or (-5)^2 - 4(m)(-2), or 25+8m.  For which values of m is 25+8m<0?  Subtracting 25 from both sides, we get 8m < -25, or m<25/8.

Let's check this.  Let m=3 (which is less than 25/8).  Does the curve ever touch the x-axis?  y = 3x^2 - 5x - 2; here a=3, b= -5 and c= -2.  Then the discriminant is 25-4(3)(-2) = 49 (which is positive), so there would be two different, real roots (x-intercepts).  No good.

Looking at y=mx^2 - 5x - 2 once more, we see that if x=0, y = -2, so for any m, the curve goes thru the point (0,-2), so long as m is positive.

Now let's look at the possibilities for negative m.  Suppose we have
y = -3x^2 - 5x - 2.  Then the discriminant would be 25-4(-3)(-2), or 1.  So we'd have 2 real, unequal roots yet again.

As a last resort, I used my calculator repeatedly to graph y=mx^2 - 5x - 2.  I found that if m is less than about -3.125, there is/are no x-intercept.   (answer)