Hint Drag to pan Mouse wheel to zoom Graph of the parabola in vertex form The vertex form of parabola equation is y = a (x h)^2 k, where (h, k) = vertex and axis of symmetry x = h The parabola is f (x) = y = 2x2 4x 3 Write the equation in vertex form of a parabola eqaution Divide each side by negative 2 y/2 = x2 2x 3 /2To graph a linear equation in slopeintercept form, we can use the information given by that form For example, y=2x3 tells us that the slope of the line is 2 and the yintercept is at (0,3) This gives us one point the line goes through, and the direction
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Y=x^2+2x-3 in graphing form
Y=x^2+2x-3 in graphing form-Up What is the axis of symmetry and concavity?We are given the linear function {eq}y=\frac{3}{2}x4 {/eq} We want to graph the given linear function So, we have Solution The given linear function is in the slopeintercept form {eq}y
How Do You Graph Y X 2 2x 3 Socratic
The inequalities are 2x 3y ≥ 6, x y ≥ 3 and y ≤ 2 Rewrite the inequalities so that solves for y , That's the slope intercept form and it will make the boundary line easier to graph 1) Draw the coordinate plane The first inequality 2x 3y ≥ 6 3y ≥ 2x 6 y ≥ ( 2/3)x 2 2) Graph the line y = ( 2/3)x 2 3) Since the inequality symbol is ≥ the boundary is \(2 x3 y=6, \quad 3 x=4 y7, \quad y=2 x5, \quad 2 y=3, \quad \text { and } \quad x2=0\) A line is completely determined by two points Therefore, to graph a linear equation, we need to find the coordinates of two points This can be accomplished by choosing an arbitrary value for x or y and then solving for the other variableAfter you enter the expression, Algebra Calculator will graph the equation y=2x1 Here are more examples of how to graph equations in Algebra Calculator Feel free to try them now Graph y=x^22x y=x^22x Graph y= (x3)^2 y= (x3)^2
Y = x 2 2x 3 is a parabola Since the x 2 is positive, it opens upward (concaveup) If you factor the right hand side, you get (x1) (x3) so that means that the xintercepts are at 1 and 3 The vertex is halfway between these of course (at x = 1) That means the vertex is at x = 1 and y = 1 2Y = x²4x3 (0,3); In this function, x = 1 would make y = 8/0 which is undefined To find this, just set the denominator equal to zero and solve for x 2x 2 = 0 2x = 2 x = 1 The function is defined for all other numbers, so the domain is (infinity, 1)U(1, infinity) I can't remember off hand any good way to find the range other than graphing the function
Changing "c" only changes the vertical position of the graph, not it's shape The parabola y = x 2 2 is raised two units above the graph y = x 2 Similarly, the graph of y = x 2 3 is 3 units below the graph of y = x 2 The constant term "c" has the same effect for any value of a and b Parabolas in the vertexform or the ahk form, y = a(x y=(x1)^216 >"the equation of a parabola in "color(blue)"vertex form" is •color(white)(x)y=a(xh)^2k "where "(h,k)" are the coordinates of the vertex and a" "is a multiplier" "to obtain this form "color(blue)"complete the square" y=x^22(1)x color(red)(1)color(red)(1)15 y=(x1)^216larrcolor(red)"in vertex form" To graph this, we'll need y y intercepts We find these just like we found x x intercepts in the previous couple of problems y 2 − 6 y 5 = 0 ( y − 5) ( y − 1) = 0 y 2 − 6 y 5 = 0 ( y − 5) ( y − 1) = 0 The parabola will have y y intercepts at y = 1 y = 1 and y = 5 y = 5 Here's a sketch of the graph
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Let us put a circle of radius 5 on a graph x 2 y 2 − 2x − 4y − 4 = 0 It is a circle equation, but "in disguise"!60 Chapter 2 Graphing and Writing Linear Equations 23 Lesson Lesson Tutorials Key Vocabulary xintercept, p 60 yintercept, p 60 slopeintercept form, p 60 Intercepts The xintercept of a line is the xcoordinate of the point where the line crosses the xaxisIt occurs when y = 0 The yintercept of a line is the ycoordinate of the point where the line crosses the yaxisAnd Standard Form Learn vocabulary, terms, and more with flashcards, games, and other study tools
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Y = a(xh)^2 k is the general form of a parabola opening Up or Down, a = 1 > 0, parabola opens Up 3rd Step what is the Line of symmetry x = 1For x = 2, y = 2 2 0 for x = 3, y = 3 2 = 5 and we obtain the solutions (0,2), (3,1), (2,0), and (3,5) which can be displayed in a tabular form as shown below If we graph the points determined by these ordered pairs and pass a straight line through them, we obtain the graph of all solutions of y = x 2, as shown in Figure 73The slopeintercept form of the equation of a line is a useful form for graphing as well as for understanding the relationship between x and y In this lesson, learn how the slopeintercept form
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23 Graphing SlopeIntercept Form Objective Give the equation of a line with a known slope and yintercept When graphing a line we found one method we could use is to make a table of values However, if we can identify some properties of the line, we may be able to y=2x 5 2) y= − 6x 4 3) y= x − 4 4) y= − x − 2Y=3x1 Polynomials Algebra Calculator can simplify polynomials, but it only supports polynomials containing the variable x Here are some examplesWrite the quadratic equation in standard form and determine if the graph opens up or down 1) y = 2x2 x – 1 2) y = 3 – x – x2 3) y = 3x2 1 – 4x 4) y = 4 – 3x 2 5) y = x 9x2 6) y = 3x 5x2–3x2 Vertex the lowest or the highest point of the graph Axis of Symmetry the vertical line through the vertex
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Rewrite in \(y=a(x−h)^{2}k\) form and determine the vertex \(y=2x^{2}−4x8\) Solution Since a = 2, factor this out of the first two terms in order to complete the square Leave room inside the parentheses to add a constant term Now use −2 to determine the value that completes the square In this case, \(\frac{(2)^{2}}{2}=\((1)^{2}=1\)Graph y=2x^2x3 y = 2x2 x − 3 y = 2 x 2 x 3 Find the properties of the given parabola Tap for more steps Rewrite the equation in vertex form Tap for more steps Complete the square for 2 x 2 x − 3 2 x 2 x 3Slopeintercept form Slope yintercept graph 1) Y = 5x 6 2) Y = ¾ x 1 3) Y = 1/4 x 3 4) Y = 2x 5 5) Y = 3/2 x 1 6) Y = 6x 1 Finding equations from graphs For each graph, find the slope, yintercept, the equation of the line that would be drawn through the points on each graph and three points on the line 1
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