Problem 180
Question
In the following exercises, graph by using intercepts, the vertex, and the axis of symmetry. $$ y=x^{2}+4 x-12 $$
Step-by-Step Solution
Verified Answer
x-intercepts: -6, 2; y-intercept: -12; vertex: (-2, -16); axis of symmetry: x = -2.
1Step 1: Identify the coefficients
For the quadratic equation written in the form y = ax^2 + bx + cidentify the coefficients a, b, and c. For the given equation, y = x^2 + 4x - 12, we have:a = 1b = 4c = -12.
2Step 2: Find the x-intercepts
To find the x-intercepts, set y = 0 and solve the quadratic equation x^2 + 4x - 12 = 0by factoring or using the quadratic formula. Factoring gives (x + 6)(x - 2) = 0So the x-intercepts are x = -6 and x = 2.
3Step 3: Find the y-intercept
The y-intercept is found by setting x = 0 in the equation y = x^2 + 4x - 12:y = 0^2 + 4(0) - 12 = -12So the y-intercept is y = -12.
4Step 4: Find the vertex
The x-coordinate of the vertex of a parabola y = ax^2 + bx + cis given by the formula x = -b/(2a). For the given equation, a = 1 and b = 4, so:x = -4/(2*1) = -2Substitute x = -2 back into the original equation to find the y-coordinate:y = (-2)^2 + 4(-2) - 12 = 4 - 8 - 12 = -16So the vertex is at (-2, -16).
5Step 5: Determine the axis of symmetry
The axis of symmetry of the parabola is the vertical line that passes through the vertex. Therefore, it is x = -2.
6Step 6: Plot the points and sketch the graph
Plot the x-intercepts (-6, 0) and (2, 0), the y-intercept (0, -12), and the vertex (-2, -16) on the coordinate plane. Draw the axis of symmetry as a vertical line through x = -2. Sketch the parabola opening upwards, passing through these points.
Key Concepts
Quadratic EquationsX-InterceptsY-InterceptVertexAxis of Symmetry
Quadratic Equations
Quadratic equations are polynomials of the form \(y = ax^2 + bx + c\). They are important because they represent parabolas when graphed. The general shape of a parabola can be upwards or downwards depending on the coefficient \(a\). If \(a > 0\), the parabola opens upwards, and if \(a < 0\), it opens downwards.
The quadratic equation for the given exercise is \(y = x^2 + 4x - 12\). Here, \(a = 1\), \(b = 4\), and \(c = -12\). Understanding these coefficients helps in finding the x-intercepts, y-intercept, vertex, and axis of symmetry.
The quadratic equation for the given exercise is \(y = x^2 + 4x - 12\). Here, \(a = 1\), \(b = 4\), and \(c = -12\). Understanding these coefficients helps in finding the x-intercepts, y-intercept, vertex, and axis of symmetry.
X-Intercepts
X-intercepts are the points where the graph crosses the x-axis. At these points, \(y = 0\). To find the x-intercepts of a quadratic equation, we need to solve the equation \(ax^2 + bx + c = 0\).
For \(y = x^2 + 4x - 12\), we set \(y = 0\) to get \(x^2 + 4x - 12 = 0\). This can be factored as \((x + 6)(x - 2) = 0\). Solving for \(x\) gives us the x-intercepts: \(x = -6\) and \(x = 2\). These points are where the graph will cross the x-axis.
For \(y = x^2 + 4x - 12\), we set \(y = 0\) to get \(x^2 + 4x - 12 = 0\). This can be factored as \((x + 6)(x - 2) = 0\). Solving for \(x\) gives us the x-intercepts: \(x = -6\) and \(x = 2\). These points are where the graph will cross the x-axis.
Y-Intercept
The y-intercept is the point where the graph crosses the y-axis. At this point, \(x = 0\).
To find the y-intercept, we substitute \(x = 0\) in the quadratic equation \(y = x^2 + 4x - 12\). This gives us:
\(y = 0^2 + 4(0) - 12 = -12\).
So, the y-intercept is \(y = -12\), and the point on the graph will be \((0, -12)\).
To find the y-intercept, we substitute \(x = 0\) in the quadratic equation \(y = x^2 + 4x - 12\). This gives us:
\(y = 0^2 + 4(0) - 12 = -12\).
So, the y-intercept is \(y = -12\), and the point on the graph will be \((0, -12)\).
Vertex
The vertex of a parabola is its highest or lowest point, and it represents the maximum or minimum value of the quadratic equation. The x-coordinate of the vertex can be found using the formula \(x = -\frac{b}{2a}\).
For the given quadratic equation \(y = x^2 + 4x - 12\), \(a = 1\) and \(b = 4\), so:
\(x = -\frac{4}{2 \cdot 1} = -2\).
To find the y-coordinate, substitute \(x = -2\) back into the equation:
\(y = (-2)^2 + 4(-2) - 12 = 4 - 8 - 12 = -16\).
The vertex is at \((-2, -16)\).
For the given quadratic equation \(y = x^2 + 4x - 12\), \(a = 1\) and \(b = 4\), so:
\(x = -\frac{4}{2 \cdot 1} = -2\).
To find the y-coordinate, substitute \(x = -2\) back into the equation:
\(y = (-2)^2 + 4(-2) - 12 = 4 - 8 - 12 = -16\).
The vertex is at \((-2, -16)\).
Axis of Symmetry
The axis of symmetry is a vertical line that passes through the vertex of the parabola and divides it into two symmetrical halves. For the quadratic equation \(y = ax^2 + bx + c\), the axis of symmetry can be found using the x-coordinate of the vertex, which is \(x = -\frac{b}{2a}\).
In our example, the axis of symmetry will be the vertical line passing through \(x = -2\). This means the equation of the line of symmetry is \(x = -2\).
To visualize this, draw a vertical line through the point \(x = -2\) on your graph.
In our example, the axis of symmetry will be the vertical line passing through \(x = -2\). This means the equation of the line of symmetry is \(x = -2\).
To visualize this, draw a vertical line through the point \(x = -2\) on your graph.
Other exercises in this chapter
Problem 178
In the following exercises, find the \(x\) - and \(y\) -intercepts. $$ y=-x^{2}-14 x-49 $$
View solution Problem 179
In the following exercises, graph by using intercepts, the vertex, and the axis of symmetry. $$ y=x^{2}+6 x+5 $$
View solution Problem 182
In the following exercises, graph by using intercepts, the vertex, and the axis of symmetry. $$ y=x^{2}-6 x+8 $$
View solution Problem 184
In the following exercises, graph by using intercepts, the vertex, and the axis of symmetry. $$ y=-x^{2}+8 x-16 $$
View solution