Question

explain why the graph of the quadratic function (f(x)=x^{2}+x + 1) 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 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.

b. 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).

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 (y = 1) because that is the solution to the equation (f(y)=0).

d. 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 = 1) because that is the solution to the equation (f(0)=0).

Explanation:

Step1: Recall x - and y - intercept concepts

The y - intercept of a function \(y = f(x)\) is found by setting \(x = 0\). For \(f(x)=x^{2}+x + 1\), when \(x = 0\), \(f(0)=0^{2}+0 + 1=1\), so the graph crosses the y - axis at \(y = 1\).

Step2: Analyze x - intercepts

The x - intercepts are found by solving \(f(x)=0\), i.e., \(x^{2}+x + 1 = 0\). Use the quadratic formula \(x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\) for a quadratic equation \(ax^{2}+bx + c = 0\). Here, \(a = 1\), \(b = 1\), \(c = 1\). Then the discriminant \(\Delta=b^{2}-4ac=1^{2}-4\times1\times1=1 - 4=- 3<0\). When \(\Delta<0\), there are no real solutions to the quadratic equation, so the graph does not cross the x - axis.

Answer:

D. 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 \(y = 1\) because \(f(0)=1\).