One more Quadratic function question

[sparklemetink]

State the conditions of the values a, b, and c so that the quadratic formula f(x)=ax^2+bx+c has no x-intercepts.

I don't know how to set this up.

 

[mmm4444bot]

Are you familiar with the Discriminant in the Quadratic Formula ?

b^2 - 4ac

It's the expression inside the radical sign (i.e., the radicand).

We call this radicand the Discriminant because its value distinguishes for us the number of distinct Real solutions to the equation f(x) = 0.

Of course, you already understand that there is an x-intercept everywhere f(x) = 0 because f(x) is the same as y, yes?

If b^2 - 4ac represents a positive value, there are two distinct Real solutions (i.e., two x-intercepts).

If b^2 - 4ac represents zero, there is one distinct Real solution (i.e., one x-intercept).

If b^2 - 4ac represents a negative value, there are no Real solutions (i.e., no x-intercepts).

You can study more about the Discriminant HERE.

 

[soroban]

Hello, sparklemetink1

mmm444bot has the best approach.
Here's another . . .

\(\displaystyle \text{State the conditions of the values }a, b, c\text{ so that the quadratic equation }y \:=\:ax^2+bx+c\text{ has no }x\text{-intercepts.}\)

\(\displaystyle \text{Assume that }a\,>\,0 \;\;\;\hdots\;\;\;\text{The parabola opens upward.}\)


\(\displaystyle \text{We know that the vertex of the parabola is at: }\:x \:=\:\frac{\text{-}b}{2a}\)

\(\displaystyle \text{The }y\text{-coordinate of the vertex is: }\:y \;=\;a\left(\frac{\text{-}b}{2a}\right)^2 + b\left(\frac{\text{-}b}{2a}\right) + c \;=\;c - \frac{b^2}{4a}\)

\(\displaystyle \text{The vertex is at: }\;\left(\frac{\text{-}b}{2a},\; c - \frac{b^2}{4a}\right)\)


\(\displaystyle \text{If the parabola has no }x\text{-intercepts, the vertex is }above\text{ the }x\text{-ais.}\)

\(\displaystyle \text{Therefore: }\;c - \frac{b^2}{4a} \:>\:0 \quad\Rightarrow\quad c\;>\:\frac{b^2}{4a} \quad\Rightarrow\quad 4ac \:>\:b^2 \quad\Rightarrow\quad b^2-4ac\:<\:0\)

. . I hope that looks familiar.


I'll let you prove the case when \(\displaystyle a \,<\,0\)

 

[sparklemetink]

Thank you both for your answers. I had forgotten b^2-4ac was called the Discriminant. I remember this now. This all looks familiar and I will let you know if I have any problems.

Thanks

 

[Tinkermom]

Ok, I don't understand how you got the y-coordinate of the vertex. I understand you put the x-coordinate into ax^2+bx+c, but I don't see how you came to the c-b^2/4a.

:/

 

[Aladdin]

Ok, I don't understand how you got the y-coordinate of the vertex. I understand you put the x-coordinate into ax^2+bx+c, but I don't see how you came to the c-b^2/4a.

:/

\(\displaystyle \text{The }y\text{-coordinate of the vertex is: }\:y \;=\;a\left(\frac{\text{-}b}{2a}\right)^2 + b\left(\frac{\text{-}b}{2a}\right) + c \;= \frac{-ab^2}{4a^2} - \frac{b^2}{2a} + c

Do you know how to add fractions ?\)