SOLUTION: Find an equation for a Parabola with vertex (1,0) containing the point(4,18) If lines 2x-y=17 and 5x+y=11 are graphed what is their intersection point? If f(x)=x2-x, what is f(-1

Algebra ->  Algebra  -> Quadratic-relations-and-conic-sections -> SOLUTION: Find an equation for a Parabola with vertex (1,0) containing the point(4,18) If lines 2x-y=17 and 5x+y=11 are graphed what is their intersection point? If f(x)=x2-x, what is f(-1      Log On

 Ad: Algebra Solved!™: algebra software solves algebra homework problems with step-by-step help! Ad: Algebrator™ solves your algebra problems and provides step-by-step explanations!

 Algebra: Conic sections - ellipse, parabola, hyperbola Solvers Lessons Answers archive Quiz In Depth

 Click here to see ALL problems on Quadratic-relations-and-conic-sections Question 52351: Find an equation for a Parabola with vertex (1,0) containing the point(4,18) If lines 2x-y=17 and 5x+y=11 are graphed what is their intersection point? If f(x)=x2-x, what is f(-1) Find an equation that represents a Parabola on the xy-plane with vertex(-3,2)opening upwards?Found 2 solutions by AnlytcPhil, funmath:Answer by AnlytcPhil(1276)   (Show Source): You can put this solution on YOUR website!```Find an equation for a Parabola with vertex (1,0) containing the point(4,18) To do this you must know that 1. The equation of a parabola with vertex (h,k) is y = a(x-h)² + k 2. If h, k, and a point (x1,y1) are known for a parabola with the equation above in 1, then the value of "a" can be found by substituting x1 for x and y1 for y and then solving for "a". So for your problrm (h,k) = (1,0), and (x1,y1) = (4,18), so we substitute 1 for h, 0 for k, 4 for x and 18 for y in y = a(x-h)² + k 18 = a(4-1)² + 0 18 = a(3)² 18 = 9a 2 = a Now rewrite the equation y = a(x-h)² + k with 2 for a, 1 for h and 0 for k, leaving x and y as variables: y = 2(x-1)² + 0 y = 2(x-1)² and if you like you may multiply that out and get y = 2x² - 4x + 2 It's graph is =================================================================== If lines 2x - y = 17 and 5x + y = 11 are graphed what is their intersection point? Write them one under the other and add them vertically term-by-term and the y's will cancel: 2x - y = 17 5x + y = 11 ----------- 7x = 28 x = 4 Substitute 4 for x in either equation I'll arbitrarily pick the second one: 5x + y = 11 5(4) + y = 11 20 + y = 11 y = -9 So they intersect at the point (4, -9) Here are their graphs: ============================================= If f(x)=x2-x, what is f(-1) Just substitute -1 for x in f(x) = x² - x and simplify the right side f(-1) = (-1)² - (-1) f(-1) = 1 + 1 f(-1) = 2 ================================================= Find an equation that represents a Parabola on the xy-plane with vertex(-3,2)opening upwards? Like the first problem we know it has equation y = a(x - h)² + k with h = -3 and k = 2. We also need to know that if "a" is a positive number the parabola will open upward and if "a" is a negative number it will open downward. ("a" may not be 0). Since it doesn't matter what "a" equals, as long as it is positive, we'll just choose "a" to be the easiest positive value we can think of, which is 1. y = 1[x - (-3)]² + 2 y = 1(x + 3)² + 2 y = (x + 3)² + 2 y = x² + 6x + 9 + 2 y = x² + 6x + 11 Here is its graph Edwin``` Answer by funmath(2925)   (Show Source): You can put this solution on YOUR website!For the first one, you need to know the vertex form of the equation of a parabola. y=a(x-h)^2+k The vertex is (h,k) notice that the h seems to change signs. Your vertex is: (1,0) y=a(x-1)^2+0 y=a(x-1)^2 Substitute the point (4,18) in for (x,y) and solve for a. 18=a(4-1)^2 18=a(3)^2 18=9a 18/9=9a/9 2=a Substitute the a into the equation and you have: y=2(x-1)^2 For the second one,using the substitution method is easiest: Solve the second equation for y and substitute it into the first equation where the y is. Solve for x. Substitute the x value you get into the second equation and you'll get y. 5x+y=11 -5x+5x+y=-5x+11 y=-5x+11 Sustitute into first line: 2x-(-5x+11)=17 2x+5x-11=17 7x-11=17 7x-11+11=17+11 7x=28 7x/7=28/7 x=4 Substitute into the second line: 5(4)+y=11 20+y=11 -20+20+y=-20+11 y=-9 The intersection point (x,y)=(4,-9) For the third one, I am going to assume that x2 is really x squared or x^2. Substitute -1 in for x: f(x)=x^2-1 f(-1)=(-1)^2-1 f(-1)=(-1)(-1)-1 f(-1)=1-1 f(-1)=0 For the forth one use the vertex form from the first problem you asked: y=(x-(-3))^2+(2) y=(x+3)^2+2