- #1

songoku

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- 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

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