Question

Help Please! Medal!

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

Answer 1

"m" does not mean "slope" in a quadratic equation. It is just a parameter.

1) m>0 or it will ALWAYS have intercepts.
2) x = 5/(2m) is the x-location of the minimum value.
3) Can you evaluate the function at x = 5/(2m) and observe the result.

Answer 2

Here are the options:
\[m>-\frac{ 25 }{ 8 }\]
\[m<-\frac{ 25 }{ 8 }\]
\[m<\frac{ 25 }{ 8 }\]
\[m>\frac{ 25 }{ 8 }\]

Answer 3

Okay, follow the steps I provided.

Answer 4

Sorry for the mistake there @Catseyeglint911

Answer 5

So since m needs to be greater than zero, the best answer would be the last one. correct?

...and no worries. it's fine. ^^ @Cosmichaotic

Answer 6

@tkhunny

Answer 7

Okay, easier way.

Can you solve: mx^2 - 5x - 2 = 0 using the Quadratic Formula?

Answer 8

BTW, my #1 was also wrong. We can have negative values. Hearty "good work" to @Cosmichaotic for making me think about it more.

Answer 9

I don't think so because you don't know the value of m, and ah okay.

Answer 10

Not so! That is the key to the problem!

\(b^{2} - 4ac \rightarrow (-5)^{2} - 4(m)(-2) = 25 + 8m\)

I hope you recognize the discriminant from the Quadratic Formula.

The deal is this, if \(25 + 8m \ge 0\), then the ARE x-intercepts.

For what values of \(m\;is\;25 + 8m \lt 0\)? THAT is the question.

Answer 11

-25/8 < 0

Answer 12

Because m has to be a negative value that is greater than 25/8 in order to make that side of the inequality less than zero.

Answer 13

Right?