How do you graph the parabola #y= 5/4(x+2)^2 -1# using vertex, intercepts and additional points?

Answer 1

Vertex: #(-2,-1)#

Y-intercept: #(0,5)#

X-intercepts: #(2/(sqrt5)-2,0)##(-2/(sqrt5)-2,0)#

Graph:

#y=5/4(x+2)^2-1# is a quadratic equation in vertex form:
#y=a(x-h)^2+k#,

where:

#a=5/4#, #h=-2#, and #k=-1#
Vertex: maximum or minimum point of the parabola, #(h,k)#
#(-2,-1)#
Y-intercept: value of #y# when #x=0#
#y=5/4(0+2)^2-1#
#y=5/4(4)-1#
Cancel #4/4#.
#y=5-1#
#y=4#
y-intercept: #(0,4)#
X-intercepts: values of #x# when #y=0#

Note: This is a long process.

Substitute #0# for #y#.
#0=5/4(x+2)^2-1#

Simplify.

#0=(5(x+2)^2)/4-1#
Add #1# to both sides.
#1=(5(x+2)^2)/4#
Multiply both sides by #4#.
#4xx1=(5(x+2)^2)/4xx4#
Cancel #4# on right-hand side.
#4=5(x+2)^2#
Divide both sides by #5#.
#4/5=(5(x+2)^2)/5#
Cancel #5# on the right-hand side.
#4/5=(x+2)^2#

Take the square root of both sides.

#+-sqrt(4/5)=x+2#

Simplify.

#+-(sqrt4)/(sqrt5)=x+2#

Simplify.

#+-2/(sqrt5)=x+2#
Subtract #2# from both sides.
#-2+-2/(sqrt5)=x#

Switch sides.

#x=+-2/(sqrt5)-2#
x-intercepts: #(2/(sqrt5)-2,0)##(-2/(sqrt5)-2,0)#
Approximate x-intercepts: #(1.106,0)##(-2.894,0)#

Summary:

Vertex: #(-2,-1)#
Y-intercept: #(0,5)#
X-intercepts: #(2/(sqrt5)-2,0)##(-2/(sqrt5)-2,0)#
Approx. x-intercepts: #(1.106,0)##(-2.894,0)#

Plot the points and sketch a parabola through the points. Do not connect the dots.

graph{y=5/4(x+2)^2-1 [-10, 10, -5, 5]}

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Answer 2

To graph the parabola (y = \frac{5}{4}(x + 2)^2 - 1), follow these steps:

  1. Vertex: The vertex form of a parabola is (y = a(x - h)^2 + k), where ((h, k)) is the vertex. In this case, the vertex is ((-2, -1)).

  2. Intercepts:

    • y-intercept: Set (x = 0) and solve for (y). (y = \frac{5}{4}(0 + 2)^2 - 1 = \frac{5}{4}(4) - 1 = 5 - 1 = 4). So, the y-intercept is ((0, 4)).
    • x-intercept: Set (y = 0) and solve for (x). (0 = \frac{5}{4}(x + 2)^2 - 1). Solve for (x) to find the x-intercepts.
  3. Additional Points: Choose additional values of (x) to calculate corresponding (y) values. For instance, choose (x = -3, -1, 1) and find the corresponding (y) values.

  4. Plot the vertex, intercepts, and additional points on the coordinate plane. Draw the parabola passing through these points.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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