# Find the range of values of a + b

• songoku
I understand now.In summary, the range of values of ##a+b## when ##y=-\frac{1}{8}x^2+ax+b## is tangent to the x-axis is ##a+b \leq \frac{1}{8}##. This can be found by setting the derivative of the function to 0 and finding the corresponding value of ##b##, which is ##-2a^2##. Then, using the fact that ##y## must also be less than or equal to 0, we can express ##a+b## as a function of ##a## and find its minimum or maximum value, which corresponds to the upper and lower bounds of the range for ##a
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

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

songoku said:
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

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?

songoku
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 and SammyS
PeroK said:
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

songoku said:
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?

songoku
songoku said:
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

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