Find the range of values of a + b

[songoku]

Homework Statement

Given that $y=-\frac{1}{8}x^2+ax+b$, find the range of values of $a+b$ if $y$ is tangent to x-axis

Relevant Equations

Quadratic

Discriminant

Derivative

The answer is $a+b \leq \frac{1}{8}$ but I don't know how to get it.

Tangent to the x-axis means the vertex is at the x-axis so the y coordinate of the vertex = 0

$$y=-\frac{1}{8}x^2+ax+b$$
$$y'=0$$
$$-\frac{1}{4}x+a=0$$
$$x=4a \rightarrow y=2a^2+b$$

So
$$2a^2+b=0$$
$$b=-2a^2$$

$y$ will also satisfy $y \leq 0$ so
$$-\frac{1}{8}x^2+ax+b \leq 0$$

Since $b=-2a^2$, finding restriction for $a+b$ is the same as finding restriction for $-a^2$

How to find the restriction by using all of those?

Or is it simply taking $x=1$ and put it into $-\frac{1}{8}x^2+ax+b \leq 0$ and the reason is "because it works"?

Thanks

 

[PeroK]

Homework Statement:: Given that $y=-\frac{1}{8}x^2+ax+b$, find the range of values of $a+b$ if $y$ is tangent to x-axis
Relevant Equations:: Quadratic

Discriminant

Derivative

The answer is $a+b \leq \frac{1}{8}$ but I don't know how to get it.

Tangent to the x-axis means the vertex is at the x-axis so the y coordinate of the vertex = 0

$$y=-\frac{1}{8}x^2+ax+b$$
$$y'=0$$
$$-\frac{1}{4}x+a=0$$
$$x=4a \rightarrow y=2a^2+b$$

So
$$2a^2+b=0$$
$$b=-2a^2$$

$y$ will also satisfy $y \leq 0$ so
$$-\frac{1}{8}x^2+ax+b \leq 0$$

Since $b=-2a^2$, finding restriction for $a+b$ is the same as finding restriction for $-a^2$

It's fine to here. Now you can express $a + b$ as a function of $a$. You're looking for the ????? of that function?

 

[pasmith]

You can get immediately to a^2 + \frac12b = 0 by observing that the only way a quadratic can be tangent to the x axis is if it has a double root, so that the discriminant (which here is a^2 + \frac12 b) is zero.

 

[songoku]

It's fine to here. Now you can express $a + b$ as a function of $a$. You're looking for the ????? of that function?

Sorry I don't understand your hint.

$a+b=a-2a^2$

Since the question is asking about range of $a+b$, I need to find the upper and lower bound of $a-2a^2$ ? Is this what you mean?

Thanks

 

[SammyS]

Sorry I don't understand your hint.

$a+b=a-2a^2$

Since the question is asking about range of $a+b$, I need to find the upper and lower bound of $a-2a^2$ ? Is this what you mean?

Thanks

What is the minimum or maximum value which $a-2a^2$ can have?

 

[PeroK]

Sorry I don't understand your hint.

$a+b=a-2a^2$

Since the question is asking about range of $a+b$, I need to find the upper and lower bound of $a-2a^2$ ? Is this what you mean?

Thanks

Yes, exactly. You need to find the range of the function $a - 2a^2$, which is another quadratic

 

[songoku]

Thank you very much for the help PeroK, pasmith, SammyS