How do you graph the function, label the vertex, axis of symmetry, and x-intercepts. #x-4 = 1/4 (y+1)^2#?

Answer 1

The vertex is at (#4,-1#).
The axis of symmetry is #y = -1#.
The #x#-intercept is (#17/4,0#).
There are no #y#-intercepts.

The standard form for the equation of a parabola is

#y= a^2 + bx +c#

Your equation is

#x−4 = 1/4(y+1)^2#

We must get this into standard form.

#x−4=1/4(y^2 + 2y + 1)#

#x−4=1/4y^2 + 1/4(2y)+ 1/4#

#x−4=1/4y^2 + 1/2y + 1/4#

#x=1/4y^2 + 1/2y + 1/4 + 4#

#x=1/4y^2 + 1/2y + 17/4#

This is standard form, but with #x# and #y# interchanged.

We are going to get a sideways parabola.

#a = 1/4#, #b = 1/2#, and #c = 17/4#.

Vertex

Since #a > 0#, the parabola opens to the right.

The #y#-coordinate of the vertex is at

#y = –b/(2a) = -(1/2)/(2×(1/4)) = -(1/2)/(1/2) = -1#.

Insert this value of #x# back into the equation.

#x=1/4y^2 + 1/2y + 17/4#

#x=1/4(-1)^2 + 1/2(-1) + 17/4 = 1/4×1 -1/2 +17/4#

#x = 1/4 -1/2 + 17/4 = (1 -2 +17)/4 = 16/4 = 4#

The vertex is at (#4, -1#).

Axis of symmetry

The axis of symmetry must pass through the vertex, so

The axis of symmetry is #y = -1#.

#x#-intercept

To find the #x#-intercept, we set #y = 0# and solve for #x#.

#x=1/4y^2 + 1/2y + 17/4 = 1/4×0^2 + 1/2×0 + 17/4 = 0 + 0 + 17/4 = 17/4#

The #x#-intercept is at (#17/4,0#).

#y#-intercepts

To find the #y#-intercepts, we set #x = 0# and solve for #y#.

#x=1/4y^2 + 1/2y + 17/4#

#0=1/4y^2 + 1/2y + 17/4#

#0=y^2 + 2y + 17#

#y = (-b ± sqrt(b^2-4ac))/(2a)#

The discriminant #D = b^2 – 4ac = 4^2 – 4×1×17= 8 – 68 = -60#

Since #D < 0#, there are no real roots.

There are no #y#-intercepts.

Graph

Now we prepare a table of #x# and #y# values.

The axis of symmetry passes through #y = -1#.

Let's prepare a table with points that are 5 units on either side of the axis, that is, from #y = -6# to #y = 4#.

Plot these points.

And we have our graph. The red line is the axis of symmetry.

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Answer 2

To graph the function (x-4 = \frac{1}{4}(y+1)^2):

  1. Rewrite the equation in vertex form: (y = a(x-h)^2 + k), where ((h, k)) is the vertex.
  2. Identify the vertex: ((h, k) = (4, -1)).
  3. Plot the vertex.
  4. The axis of symmetry is (x = 4).
  5. Since (a = \frac{1}{4}), the parabola opens upwards.
  6. Determine the x-intercepts by setting (y = 0) and solving for (x).
  7. Plot the x-intercepts.
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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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