Algebra Graph y= (x1)^24 y = (x − 1)2 − 4 y = ( x 1) 2 4 Find the properties of the given parabola Tap for more steps Use the vertex form, y = a ( x − h) 2 k y = a ( x h) 2 k, to determine the values of a a, h h, and k k a = 1 a = 1 h = 1 h = 1 k = − 4 k = 4The graph of y = x 2 is a parabola that opens upwards The vertex of this parabola is at the point where x = 0 y = x 2 If you change x to x 5 so the equation is y = (x 5) 2 then the vertex is at the point where x 5 = 0, that is at x = 5 Thus a change from x to x 5 is a translation of 5 units to the right y = (x 5)2Thus the graph doesn't cross the xaxis at 2
Draw The Graph Of The Follwing Equations 2x Y 2 0 4x 3y 24 0 Y 4 0 Obtain The Vertices Of The Brainly In
Graph of y=1/2(x+4)^2
Graph of y=1/2(x+4)^2-\(y=x^2(x^21)(x^22)^4\) In each case, you should pay particular attention to the points where the curve meets the \(x\) axis To sketch the graph of (i), first notice that we must have \(x(x1)(x2)^4 \geq 0\) in order for \(y\) to take real valuesSubtract y from both sides x^ {2}4x4y=0 x 2 4 x 4 − y = 0 This equation is in standard form ax^ {2}bxc=0 Substitute 1 for a, 4 for b, and 4y for c in the quadratic formula, \frac {b±\sqrt {b^ {2}4ac}} {2a} This equation is in standard form a x 2 b x c = 0 Substitute 1 for a, 4 for b, and 4 − y for c in the quadratic
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Y=(x1)2 4 find p, axis of symmetry, vertices, foci, directrix for this parabola Thankyou! The table for example 1 is obtained by substituting the different xvalues into y = x 2 So for example, when x = 2, then y = (2) 2 = 4 This is graphed by putting a dot on the point (2,4) We then put dots for all the other points in the table (like (4,16), then (3,9) and so onCompute answers using Wolfram's breakthrough technology & knowledgebase, relied on by millions of students & professionals For math, science, nutrition, history
star 48 /5 heart 175 profile Brainly User The vertex of the graph of y = (x 1)2 5 is ( 1,5) 1jaiz4 and 279 more users found this answer helpful heart outlinedPlease Subscribe here, thank you!!! y = a(x − h)2 k where (h,k) is the vertex of the parabola For your equation that means the vertex is (1,4) Plot this point NOTE there is a sign change for the h but not k Now we need another point The easiest to find is by letting x=0 y = 2(0 − 1)2 4 y = 2( − 1)2 4
Use the distributive property to multiply y by x 2 1 Add x to both sides Add x to both sides All equations of the form ax^ {2}bxc=0 can be solved using the quadratic formula \frac {b±\sqrt {b^ {2}4ac}} {2a} The quadratic formula gives two solutions,You can clickanddrag to move the graph around If you just clickandrelease (without moving), then the spot you clicked on will be the new center To reset the zoom to the original click on the Reset button Using "a" Values There is a slider with "a =" on it You can use "a" in your formula and then use the slider to change the value of "aA line has slope 2 and passes through point (2, 4) Find the equation of the line Solution Substitute 2 for m and (2, 4) for (x 1, y 1) in Equation (2) Thus, a line with slope 2 that passes through the point (2, 4) has the equation y = 2x 8 We could also write the equation in equivalent forms y 2x = 8, 2x y = 8, or 2x y 8 = 0
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This problem has been solved!Algebra Graph y=4*2^x y = 4 ⋅ 2x y = 4 ⋅ 2 x Exponential functions have a horizontal asymptote The equation of the horizontal asymptote is y = 0 y = 0 Horizontal Asymptote y = 0 y = 0 the coordinate plane that passes through the 3, 6 and has a slope of four the yintercept of the line is Find the distance between the two points (1,1) ? (2,4) Enter the number that goes beneath the radical symbol
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Which Of The Following Graph Represents The Equation Y 2 3 X 1 Brainly Com
Algebra > Quadraticrelationsandconicsections> SOLUTION sketch the graph of each ellipse 1x^2/9y^2/4=1 2x^2/9y^2=1 3x^2y^2/4=1 4y^2/4x^2/25=1 5y^2/9X^2/16=1 6x^2/25y^2=1 7x^2y^2/9=1 8x^2y^2/25=1 9x^2/9y^2=1 Log OnQuestion y=(x1)2 4 find p, axis of symmetry, veritces, foci, directix for this parabola Thankyou!EXAMPLE 11 Sketch the graph of y = (x 2)4(x 1)3(x – 1) by finding its intercepts and its limits as x → and x → 0 х SOLUTION The yintercept is f(0) = (2)4( (1) and the xintercepts are found by setting y = 0 X = 2, 1, Notice that since (x 2)4 is positive, the function doesn't change sign at 2;
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Answer by stanbon(757) (Show Source) You can put this solution on YOUR website!Graph (y2)^2=4(x1) Rewrite the equation as Divide each term by and simplify Tap for more steps Divide each term in by Cancel the common factor of Graph the parabola using its properties and the selected points Graph the parabola using its properties and the selected points Direction Opens Left VertexRelated » Graph » Number Line » Examples » Our online expert tutors can answer this problem Get stepbystep solutions from expert tutors as fast as 1530 minutes
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The graph of the function y = f (x) is vertically stretched about the xaxis by a factor of 2 y x 2 4 6246 2 424 0 y = f (x) a) Determine the domain and range of the transformed function b) Explain the effect that a vertical stretch has on the domain and range of a functionAll equations of the form a x 2 b x c = 0 can be solved using the quadratic formula 2 a − b ± b 2 − 4 a c The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction x^ {2}4xy5=0 x 2 4 x − y − 5 = 0 This equation is in standard form ax^ {2}bxc=0 Substitute 1 for a, 4 for b, andConsider the given point and the new graph (− 3, 5) and y = g(x 1) − 3 Now, let us take x 1 = − 3 x = − 4 Now, let us take y = g(− 3) − 3 = 5 − 3 = 2 Hence, the corresponding location of the given point on the graph of y = g(x 1) − 3 is (− 4, 2)
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