Problem 5
Question
Does the graph of the function have x-intercepts of 4 and 5? \(y=2(x+4)(x-5)\)
Step-by-Step Solution
Verified Answer
No, the graph of function \(y=2(x+4)(x-5)\) does not have x-intercepts of 4 and 5. But it has intercepts of -4 and 5.
1Step 1: Identify the x-intercepts
The x-intercepts of a function are the values of x when y is zero. So to find the x-intercepts, set the equation \(y=2(x+4)(x-5)\) equal to 0. Solving it you get, \(0=2(x+4)(x-5)\). To make this equation true, either \(x+4=0\) or \(x-5=0\).
2Step 2: Solve for x
Solving the equations \(x+4=0\) and \(x-5=0\) for x, one finds the values are x=-4 and x=5.
3Step 3: Compare to the Suggested Intercepts
The x-intercepts found in step 2 (-4 and 5) are not the same as the given intercepts (4 and 5). So, the graph of function does not have intercepts of 4 and 5.
Key Concepts
Quadratic FunctionsSolving Quadratic EquationsGraphing Quadratic Functions
Quadratic Functions
Quadratic functions form the foundation for understanding various complex mathematical concepts. They are expressed in the standard form of \(y=ax^2+bx+c\), where \(a\), \(b\), and \(c\) represent constants with \(a\) not equal to zero.
These functions produce a graph called a parabola which opens upwards if \(a > 0\) or downwards if \(a < 0\). An important feature of parabolas is their symmetry, with a vertex being the maximum or minimum point. Quadratic functions also have up to two x-intercepts, the points where the graph crosses the x-axis, which are crucial in understanding the function’s roots.
In our exercise, the quadratic function given is \(y=2(x+4)(x-5)\). This is an expanded form, which can be very helpful as it immediately displays the solutions or x-intercepts when set to zero.
These functions produce a graph called a parabola which opens upwards if \(a > 0\) or downwards if \(a < 0\). An important feature of parabolas is their symmetry, with a vertex being the maximum or minimum point. Quadratic functions also have up to two x-intercepts, the points where the graph crosses the x-axis, which are crucial in understanding the function’s roots.
In our exercise, the quadratic function given is \(y=2(x+4)(x-5)\). This is an expanded form, which can be very helpful as it immediately displays the solutions or x-intercepts when set to zero.
Solving Quadratic Equations
Solving quadratic equations is central to many algebraic problems. An equation is termed quadratic when it has the form of \(ax^2+bx+c=0\), and it is solved to find the values of \(x\) that make the equation true. There are several methods to solve these equations, such as factoring, completing the square, using the quadratic formula, or graphing.
Factoring involves expressing the quadratic equation as a product of two binomials. If you can find two numbers that both add to \(b\) and multiply to \(ac\), you can rewrite the quadratic as \( (x+n)(x+m)=0 \), where \(n\) and \(m\) are the numbers found. Setting each factor equal to zero gives us the potential x-values.
In the exercise, the equation is factored already, and setting each factor to zero, as demonstrated in the solution, yields the x-intercepts (or roots) of the function.
Factoring involves expressing the quadratic equation as a product of two binomials. If you can find two numbers that both add to \(b\) and multiply to \(ac\), you can rewrite the quadratic as \( (x+n)(x+m)=0 \), where \(n\) and \(m\) are the numbers found. Setting each factor equal to zero gives us the potential x-values.
In the exercise, the equation is factored already, and setting each factor to zero, as demonstrated in the solution, yields the x-intercepts (or roots) of the function.
Graphing Quadratic Functions
Graphing is a powerful tool to visualize the behavior of quadratic functions. The graph, a parabola, can quickly reveal key aspects such as the direction it opens, vertex position, axis of symmetry, and x-intercepts.
The vertex can be found through the formula \(x=-\frac{b}{2a}\) if the quadratic is in standard form or by identifying the turning point of the parabola. The axis of symmetry is a vertical line that passes through the vertex, precisely dividing the parabola into two mirror images.
The x-intercepts, if they exist, are found where the graph crosses the x-axis. These intercepts correspond to the solutions of the quadratic equation when it is set to zero. A parabola can have zero, one, or two x-intercepts, reflecting the number of real roots the equation has. In the provided exercise, as we found, the graph of the function crosses the x-axis at \(x=-4\) and \(x=5\), not at 4 and 5 as initially questioned.
The vertex can be found through the formula \(x=-\frac{b}{2a}\) if the quadratic is in standard form or by identifying the turning point of the parabola. The axis of symmetry is a vertical line that passes through the vertex, precisely dividing the parabola into two mirror images.
The x-intercepts, if they exist, are found where the graph crosses the x-axis. These intercepts correspond to the solutions of the quadratic equation when it is set to zero. A parabola can have zero, one, or two x-intercepts, reflecting the number of real roots the equation has. In the provided exercise, as we found, the graph of the function crosses the x-axis at \(x=-4\) and \(x=5\), not at 4 and 5 as initially questioned.
Other exercises in this chapter
Problem 5
Match the trinomial with a correct factorization. $$ \begin{aligned} &A.)\quad (x+5)(x-4)\\\ &B.)\quad(x+4)(x+5)\\\ &C.)\quad(x-4)(x-5)\\\ &D.)\quad(x+4)(x-5) \
View solution Problem 5
Use a special product pattern to find the product. $$ (6+p)^{2} $$
View solution Problem 5
Copy the equation and fill in the blanks. \((x-3)(x+1)=x^{2}-2 x-\)______
View solution Problem 6
Identify the polynomial by degree and by the number of terms. $$ 4 w^{3}-8 w+9 $$
View solution