Problem 67
Question
For function of the form \(f(x)=\) \(a x^{2}+b x+c,\) find the discriminant, \(b^{2}-4 a c,\) and use it to determine the number of \(x\) -intercepts of the graph of \(f .\) Also determine the number of real solutions of the equation \(f(x)=0.\) $$f(x)=2 x^{2}+x+1$$
Step-by-Step Solution
Verified Answer
The discriminant for the function \(f(x)=2x^{2}+x+1\) is -7. Therefore, there are no x-intercepts and the equation \(f(x)=0\) has no real solutions.
1Step 1: Identify coefficients
The given function is \(f(x)=2x^{2}+x+1\). So \(a=2\), \(b=1\), and \(c=1\).
2Step 2: Calculate discriminant
The discriminant \(D\) of a quadratic equation is given by \(D=b^{2}-4ac\). Substituting \(a=2\), \(b=1\), and \(c=1\) into the formula, we get \(D= (1)^{2}-4*2*1 = 1 - 8 = -7\).
3Step 3: Analyze discriminant
The discriminant \(D=-7\) is less than zero. This means that the graph of the function does not intercept the x-axis, i.e., there are no real solutions to the equation \(f(x) = 0\). Hence, there are no x-intercepts for this function.
Key Concepts
Quadratic FunctionsX-intercepts of Quadratic FunctionReal Solutions of Quadratic Equations
Quadratic Functions
The general form of a quadratic function is expressed as \( f(x) = ax^2 + bx + c \), where \( a \), \( b \), and \( c \) are coefficients, and they determine the shape and position of the parabola that represents the function on a graph. The coefficient \( a \) is particularly important as it indicates the direction of the parabola opening; if \( a > 0 \), the parabola opens upwards, and if \( a < 0 \), it opens downwards.
A quadratic function is characterized by a curve known as a parabola. This curve has a vertex, which is the point where the function reaches its maximum or minimum value. In the canonical function defined for this exercise, \( f(x)=2x^2+x+1 \), the coefficient \( a \), being positive, implies that the parabola opens upwards, indicating that the vertex is a minimum point.
A quadratic function is characterized by a curve known as a parabola. This curve has a vertex, which is the point where the function reaches its maximum or minimum value. In the canonical function defined for this exercise, \( f(x)=2x^2+x+1 \), the coefficient \( a \), being positive, implies that the parabola opens upwards, indicating that the vertex is a minimum point.
X-intercepts of Quadratic Function
X-intercepts, also known as zeros or roots, of quadratic functions are the points where the graph of a quadratic function crosses the x-axis. To find the x-intercepts algebraically, we set the function \( f(x) \) equal to zero and solve the equation \( ax^2 + bx + c = 0 \).
The discriminant, defined as \( D = b^2 - 4ac \), plays a crucial role in determining the nature and number of x-intercepts. If the discriminant is positive, there are two distinct and real x-intercepts. If it's zero, the parabola touches the x-axis at exactly one point, meaning there is one real, repeated root. However, if the discriminant is negative, as in our function \( f(x) = 2x^2 + x + 1 \), this indicates there are no real x-intercepts since the parabola does not intersect the x-axis at all. Instead, the solutions are complex numbers.
The discriminant, defined as \( D = b^2 - 4ac \), plays a crucial role in determining the nature and number of x-intercepts. If the discriminant is positive, there are two distinct and real x-intercepts. If it's zero, the parabola touches the x-axis at exactly one point, meaning there is one real, repeated root. However, if the discriminant is negative, as in our function \( f(x) = 2x^2 + x + 1 \), this indicates there are no real x-intercepts since the parabola does not intersect the x-axis at all. Instead, the solutions are complex numbers.
Real Solutions of Quadratic Equations
The real solutions of a quadratic equation \( ax^2 + bx + c = 0 \) correspond to the x-intercepts of its graph. The nature of these solutions is determined by the discriminant. For a discriminant \( D \) that is greater than zero, the quadratic equation has two real and distinct solutions. If \( D = 0 \), there is exactly one real solution because the quadratic has a perfect square trinomial form.
When the discriminant is negative, the quadratic equation does not have real solutions. Instead, as we can see in our example where \( D=-7 \), the solutions are complex numbers and there are no real x-intercepts on the graph. Thus, in this case, the function \( f(x) \) does not cross the x-axis, and the parabola is situated either entirely above or below the x-axis, depending on the sign of the coefficient \( a \).
When the discriminant is negative, the quadratic equation does not have real solutions. Instead, as we can see in our example where \( D=-7 \), the solutions are complex numbers and there are no real x-intercepts on the graph. Thus, in this case, the function \( f(x) \) does not cross the x-axis, and the parabola is situated either entirely above or below the x-axis, depending on the sign of the coefficient \( a \).
Other exercises in this chapter
Problem 67
In Exercises \(67-86,\) find expressions for \((f \circ g)(x)\) and \((g \circ f)(x)\) Give the domains of \(f \circ g\) and \(g \circ f\). $$f(x)=-x^{2}+1 ; g(
View solution Problem 67
Find all solutions of the quadratic equation. Relate the solutions of the equation to the zeros of an appropriate quadratic function. $$5 x^{2}=-2 x-3$$
View solution Problem 68
Graph the function \(f(t)=t^{2}-4\) in a decimal window. Using your graph, determine the values of \(t\) for which \(f(t) \geq 0\).
View solution Problem 68
In Exercises \(67-86,\) find expressions for \((f \circ g)(x)\) and \((g \circ f)(x)\) Give the domains of \(f \circ g\) and \(g \circ f\). $$f(x)=2 x+5 ; g(x)=
View solution