Solve y = x² + 8x + 20: Finding Values Where Function is Positive

Look at the following function:

y=x2+8x+20 y=x^2+8x+20

Determine for which values of x x the following is true:

f(x)>0 f\left(x\right)>0

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Step-by-step written solution

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1

Understand the problem

Look at the following function:

y=x2+8x+20 y=x^2+8x+20

Determine for which values of x x the following is true:

f(x)>0 f\left(x\right)>0

2

Step-by-step solution

The function given is y=x2+8x+20 y = x^2 + 8x + 20 . This is a quadratic function where the coefficient of x2 x^2 (which is a=1 a = 1 ) is positive, indicating the parabola opens upwards.

Let’s calculate the vertex to find the minimum value of y y . The vertex of a parabola described by y=ax2+bx+c y = ax^2 + bx + c is found at x=b2a x = -\frac{b}{2a} .

Here, a=1 a = 1 , b=8 b = 8 . So the vertex is at:

x=82×1=4 x = -\frac{8}{2 \times 1} = -4

Substitute x=4 x = -4 into the function y=x2+8x+20 y = x^2 + 8x + 20 to calculate the minimum value of y y .

y=(4)2+8(4)+20=1632+20=4 y = (-4)^2 + 8(-4) + 20 = 16 - 32 + 20 = 4

The minimum value of the function is y=4 y = 4 at x=4 x = -4 .

Given the opening direction of the parabola and the positive minimum value, the function f(x)=x2+8x+20 f(x) = x^2 + 8x + 20 is always greater than 0.

Thus, the function is positive for all values of x x .

3

Final Answer

The function is positive for all values of x x .

Practice Quiz

Test your knowledge with interactive questions

The graph of the function below intersects the X-axis at points A and B.

The vertex of the parabola is marked at point C.

Find all values of \( x \) where \( f\left(x\right) > 0 \).

AAABBBCCCX

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