Question

explain why the graph of the quadratic function ( f(x) = x^2 + x + 5 ) crosses the ( y )-axis but does not cross the ( x )-axis.

choose the correct answer below.

a. the graph does not cross the ( x )-axis because there are no real solutions to the equation ( f(x) = 0 ). the graph does cross the ( y )-axis at ( y = 5 ) because that is the solution to the equation ( f(y) = 0 ).

b. the graph does not cross the ( x )-axis because the domain of ( f(x) ) is limited to positive values of ( x ). the graph crosses the ( y )-axis because the range of ( f(x) ) is all real numbers.

c. the graph does not cross the ( x )-axis because there are no real solutions to the equation ( f(x) = 0 ). the domain of ( f(x) ) is all real numbers, so the graph crosses the ( y )-axis at ( x = 0 ).

d. the graph does not cross ( x )-axis because it cannot be factored. the domain of ( f(x) ) is all real numbers, so the graph crosses the ( y )-axis at ( x = 0 ).

Explanation:

• For x - axis crossing: Solve \(f(x)=0\) (i.e., \(x^{2}+x + 5=0\)). The discriminant \(\Delta=b^{2}-4ac=1^{2}-4\times1\times5=1 - 20=- 19<0\), so no real roots (no x - axis crossing).
• For y - axis crossing: Substitute \(x = 0\) into \(f(x)\), \(f(0)=0^{2}+0 + 5 = 5\), so the graph crosses the y - axis at \(x = 0\) (since domain of quadratic is all real numbers, \(x = 0\) is allowed).
• Analyze options:
• Option A: Wrong, \(f(y)=0\) is incorrect, should be \(f(x)=0\) for x - axis and \(x = 0\) for y - axis.
• Option B: Wrong, domain of quadratic \(f(x)=x^{2}+x + 5\) is all real numbers, not limited to positive x.
• Option C: Correct, explains no real roots for \(f(x)=0\) (no x - axis) and crossing y - axis at \(x = 0\) (since domain is all real numbers).
• Option D: Wrong, not crossing x - axis because of discriminant (no real roots), not because it can't be factored.

Answer:

C. The graph does not cross the x - axis because there are no real solutions to the equation \(f(x)=0\). The domain of \(f(x)\) is all real numbers, so the graph crosses the y - axis at \(x = 0\)