Problem 4
Question
Find the coordinates of the vertex of the graph of \(y=2 x^{2}+4 x-2\)
Step-by-Step Solution
Verified Answer
The vertex of the graph of the function \(y=2x^{2}+4x-2\) is (-1, -4)
1Step 1: Identify a, b and c from the equation
From the equation \(y=2x^{2}+4x-2\), we identify that \(a = 2\), \(b = 4\), and \(c = -2\) respectively.
2Step 2: Calculate the x-coordinate of the vertex
The x-coordinate of the vertex h is given by \(h = -b/(2a)\). Substituting the values of a and b from step 1, we find \(h = -4/(2*2) = -1\). So, the x-coordinate of the vertex is -1.
3Step 3: Calculate the y-coordinate of the vertex
The y-coordinate of the vertex k is determined by substituting h into the equation. So, \(k = y(h) = 2*(-1)^2 + 4*(-1) - 2 = 2 - 4 - 2 = -4\). Therefore, the y-coordinate of the vertex is -4
Key Concepts
Quadratic EquationsCoordinate GeometryParabola Properties
Quadratic Equations
Quadratic equations are a fundamental part of algebra and involve expressions where the highest degree of any term is two. This means the equation will have a general form of:
Another key feature of quadratic equations is the parabola, a curve that is U-shaped and can open upwards or downwards. The sign of the coefficient \( a \) determines the direction in which the parabola opens:
- \( ax^2 + bx + c = 0 \)
Another key feature of quadratic equations is the parabola, a curve that is U-shaped and can open upwards or downwards. The sign of the coefficient \( a \) determines the direction in which the parabola opens:
- If \( a > 0 \), the parabola opens upwards.
- If \( a < 0 \), the parabola opens downwards.
Coordinate Geometry
Coordinate geometry, often referred to as analytic geometry, involves representing geometric shapes algebraically using the Cartesian coordinate system. In this system, each point has an \( (x, y) \) coordinate that defines its location in the plane.
For parabolas described by quadratic equations, we can efficiently determine important points like the vertex using coordinates. The coordinate system allows for visualization of the parabola as well as the calculation of distances, slopes, and angles.
Calculating the vertex's coordinates involves using the formula for the x-coordinate \( h = -\frac{b}{2a} \) derived from the equation \( ax^2 + bx + c \). Once \( h \) is calculated, it's substituted back into the equation to find the y-coordinate, \( k \). This effectively gives the vertex in the form \((h, k)\), which pinpoints the 'tip' of the parabola, signifying where it changes direction.
For parabolas described by quadratic equations, we can efficiently determine important points like the vertex using coordinates. The coordinate system allows for visualization of the parabola as well as the calculation of distances, slopes, and angles.
Calculating the vertex's coordinates involves using the formula for the x-coordinate \( h = -\frac{b}{2a} \) derived from the equation \( ax^2 + bx + c \). Once \( h \) is calculated, it's substituted back into the equation to find the y-coordinate, \( k \). This effectively gives the vertex in the form \((h, k)\), which pinpoints the 'tip' of the parabola, signifying where it changes direction.
Parabola Properties
Parabolas have unique geometric properties that arise from quadratic equations. They are symmetrical and can be shifted, stretched, or compressed. The vertex of a parabola is its most noticeable feature.
Some key properties include:
Some key properties include:
- Vertex: The vertex forms the turning point of the parabola, either the lowest or highest point depending on the direction it opens.
- Axis of symmetry: This is a vertical line passing through the vertex and divides the parabola into two mirror-image halves. Its equation is \( x = h \).
- Focus and Directrix: While not always needed for basic vertex calculations, the focus is a point inside the parabola that helps define its shape, and the directrix is a line outside it. Each point on the parabola is equidistant from the focus and the directrix.
Other exercises in this chapter
Problem 4
Find the discriminant for the equation. Then tell if the equation has two solutions, one solution, or no real solution. $$-3 x^{2}+6 x-3=0$$
View solution Problem 4
Use the quadratic formula to solve the equation. $$x^{2}+6 x-7=0$$
View solution Problem 5
Find the discriminant for the equation. Then tell if the equation has two solutions, one solution, or no real solution. $$x^{2}-5 x-10=0$$
View solution Problem 5
Use the quadratic formula to solve the equation. $$x^{2}-2 x-15=0$$
View solution