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How can a parabola have no x intercepts?
Suppose the equation of the parabola is y = ax2 + bx + c Now, where the parabola crosses the x-axis (the x intercepts), the value of y must be zero (that is what crossing the x-axis means). If the discriminant, b2 - 4ac is less than zero, y has no real roots. This means that there is no real value of x for which y equals zero and so the parabola has no x intercepts. If the discriminant is zero then the parabola only touches the x-axis - at (-b/2a,0). If the discriminant is greater than zero, there are two distinct intercepts. If a>0 then the parabola is shaped like a U and is wholly above the x-axis. If a<0 then the parabola is an upturned U, wholly below the x axis. If a = 0 the quadratic term disappears and the function is a straight line, not a parabola.
How do you graph quadratic functions in vertex form?
The standard form of quadratic function is: f(x) = a(x - h)^2 + k, a is different than 0 The graph of f is a parabola whose vertex it is the point (h, k). If a > 0, the parabola opens upward; if a < 0, the parabola opens downward. Furthermore, if |a| is small, the parabola opens more flatly than if |a| is large. It is a general procedure for graphing parabolas whose equations are in standard form: Example 1: Graph the the quadratic function f(x) = -2(x - 3)^2 + 8 Solution: Standard form: f(x) = a(x - h)^2 + k Given function: f(x) = -2(x - 3) + 8 From the give function we have: a= -2; h= 3; k = 8 Step 1. Determine how the parabola opens. Note that a = -2. Since a < 0, the parabola is open downward. Step 2. Find the vertex. The vertex of parabola is at (h, k). because h = 3 and k = 8, the parabola has its vertex at (3, 8). Step 3. Find the x-intercepts by solving f(x) = 0. Replace f(x) with 0 at f(x) = -2(x - 3)^2 + 8 and solve for x 0 = -2(x - 3)^2 + 8 2(x - 3)^2 = 8 (x- 3)^2 = 4 x - 3 = square radical 4 x - 3 = 2 or x -3 = -2 x = 5 or x = 1 The x- intercepts are 1 and 5. Thus the parabola passes through the points (1, 0) and (5, 0), this means that parabola intercepts the x-axis at 1 and 5. Step 4. Find the y-intercept by computing f(0). Replace x with 0 in f(x) = _2(x - 3)^2 + 8 f(0) = -2(0 - 3)^2 + 8 f(0) = -2(9) + 8 f(0) = -10 The y-intercept is -10. Thus the parabola passes through the point (0, -10), this means that parabola intercepts the y-axis at -10. Step 5. Graph the parabola. With a vertex at (3, 8), x-intercepts at 1 and 5, and a y-intercept at -10. The axis of symmetry is the vertical line whose equation is x = 3. Example 2: Graphing a quadratic function in the form f(x) = ax^2 + bx + c Graph the quadratic function f(x) = -x^2 - 2x + 1 Solution: Here a = -1, b = -2, and c = 1 Step 1. Determine how the parabola opens. Since a = 1, a < 0, the parabola opens downward. Step 2. Find the vertex. We know that x-coordinate of the vertex is x = -b/2a. Substitute a with -1 and b with -2 into the equation for the x-coordinate: x = - b/2a x= -(-2)/(2)(-1) x = -1, so the x-coordinate of the vertex is -1, and the y-coordinate of the vertex will be f(-1). thus the vertex is at ( -1, f(-1) ) f(x) = -x^2 - 2x +1 f(-1) = -(-1)^2 - 2(-1) + 1 f(-1) = -1 + 2 + 1 f(-1) = 2 So the vertex of the parabola is (-1, 2) Step 3. Find the x-intercepts by solving f(x) = o f(x) = -x^2 -2x + 1 0 = -x^2- 2x + 1 We can't solve this equation by factoring, so we use the quadratic formula to solve it. we get to solution: One solution is x = -2.4 and the other solution is 0.4 (approximately). Thus the x-intercepts are approximately -2.4 and 0.4. The parabola passes through ( -2.4, 0) and (0.4, 0) Step 4. Find the y-intercept by computing f(0). f(x) = -x^2 - 2x + 1 f(0) = -(0)^2 - 2(0) + 1 f(0) = 1 The y-intercept is 1. The parabola passes through (0, 1). Step 5. graph the parabola with vertex at (-1, 2), x-intercepts approximately at -2.4 and 0.4, and y -intercept at 1. The line of symmetry is the vertical line with equation x= -1.
Is .625 greater than 0?
Yes. 0.625 is greater than zero, but it is less than one.
What is the equation of a prabola with the vertex 0 0 and focus 0 4?
x2 = 16y The standard formula for a parabola with its vertex at the origin (0, 0) and a given focus (and the y-axis as an axis of symmetry) is as follows: x2 = 4cy In this case, the c is the y value of the focus. The focus in this case was (0, 4), and the y value in the focus is 4. That makes the c = 4. Further, that makes the equation for this parabola x2 = 4 (c)y = 4 (4)y = 16y Given that the vertex was the origin, (0, 0), and the focus is (0, 4), we can conclude that the axis of symmetry is the y-axis because the y value of the focus is 0. We can also conclude that the parabola opens up, because the focus has a positive y value.
Is four greater than negative eleven?
It is, not counting the divider 0, 15 places greater than - 11.
What way does the parabola open if a is greater than 0?
If a is greater than zero then the parabola opens upward.
When does a parabola open down?
No, a parabola is the whole curve, not just a part of it.
What direction does the parabola open?
If the equation of the parabola isy = ax^2 + bx + c, then it opens above when a>0 and opens below when a<0. [If a = 0 then the equation describes a straight line, and not a parabola!].
Why B2 - 4AC equals 0is criteria for satisfaction of parabola?
It is not clear what this question means. You can have a perfectly satisfactory parabola when b2 - 4ac is less than or greater than 0.
When will parabola open down?
In classic geometry, it opens down when the directrix is above the focus.In analytical (coordinate) geometry, if the equation of the parabola isy = ax^2 + bx + c, it opens down if a < 0.
What is the equation of a parabola with a vertex at 0 0 and a focus at 0 6?
The standard equation for a Parabola with is vertex at the origin (0,0) is, x2 = 4cy if the parabola opens vertically upwards/downwards, or y2 = 4cx when the parabola opens sideways. As the focus is at (0,6) then the focus is vertically above the vertex and we have an upward opening parabola. Note that c is the distance from the vertex to the focus and in this case has a value of 6 (a positive number). The equation is thus, x2 = 4*6y = 24y
How can a parabola have no x intercepts?
Suppose the equation of the parabola is y = ax2 + bx + c Now, where the parabola crosses the x-axis (the x intercepts), the value of y must be zero (that is what crossing the x-axis means). If the discriminant, b2 - 4ac is less than zero, y has no real roots. This means that there is no real value of x for which y equals zero and so the parabola has no x intercepts. If the discriminant is zero then the parabola only touches the x-axis - at (-b/2a,0). If the discriminant is greater than zero, there are two distinct intercepts. If a>0 then the parabola is shaped like a U and is wholly above the x-axis. If a<0 then the parabola is an upturned U, wholly below the x axis. If a = 0 the quadratic term disappears and the function is a straight line, not a parabola.
What is the vertex for the parabola y equals x squared plus 4x plus 5?
The vertex of a parabola is the minimum or maximum value of the parabola. To find the maximum/minimum of a parabola complete the square: x² + 4x + 5 = x² + 4x + 4 - 4 + 5 = (x² + 4x + 4) + (-4 + 5) = (x + 2)² + 1 As (x + 2)² is greater than or equal to 0, the minimum value (vertex) occurs when this is zero, ie (x + 2)² = 0 → x + 2 = 0 → x = -2 As (x + 2)² = 0, the minimum value is 0 + 1 = 1. Thus the vertex of the parabola is at (-2, 1).
Is 1.2 greater than or less than 0?
Greater than 0
Is 0.9 greater than 0?
Yes....0.9 is greater than 0. 0.9 > 0
Is 0 greater than -22?
Yes 0 is greater than -22
Is greater than -811.78 greater than 0?
No.
