SOLUTION: Will you please help me with the following problem? Put the following quadratic in standard form, identify the vertex, y-intercept, and graph including two symmetrical points to

Algebra ->  Quadratic Equations and Parabolas  -> Quadratic Equations Lessons  -> Quadratic Equation Lesson -> SOLUTION: Will you please help me with the following problem? Put the following quadratic in standard form, identify the vertex, y-intercept, and graph including two symmetrical points to      Log On


   

Question 591: Will you please help me with the following problem?
Put the following quadratic in standard form, identify the vertex, y-intercept, and graph including two symmetrical points to the left and right of the vertex.
Answer by janinecb(25) About Me  (Show Source):
You can put this solution on YOUR website!
Different books actually have slightly different standard forms for the parabola. One version is:
y+=+a%28x+-+h%29%5E2+%2B+k Where (h, k) is the vertex.

To find the y-intercept, just plug in zero for the x and solve for y.

To find two symmetrical points, find two numbers the same distance to the left and right of the x value of the vertex. Then plug them in for x and solve for y.

Example:
y+=+%28x+-+2%29%5E2+%2B+1 The vertex is (2, 1).

Plug in zero for x to find the y-intercept.
y+=+%280+-+2%29%5E2+%2B+1
y+=+%28-2%29%5E2+%2B+1
y = 4 + 1
y = 5 So the y-intercept in point form is (0, 5).

Pick two x values that are the same distance from the x value of the vertex, such as 1 and 3. Then plug them into the equation to find the y values to go with them on the graph.
y+=+%281+-+2%29%5E2+%2B+1
y+=+%28-1%29%5E2+%2B+1
y = 1 + 1
y = 2 So one point is (1, 2).

y+=+%283+-+2%29%5E2+%2B+1
y+=+%281%29%5E2+%2B+1
y = 1 + 1
y = 2 So a symmetrical point is (3, 2).

graph%28300%2C+200%2C+-2%2C+6%2C+-2%2C+10%2C+%28x+-+2%29%5E2+%2B1%29