Q. Which graph represents the function \(f(x)=(x-5)^2+3\)?

Answer

Shifted parabola. From the parent function \(y=x^2\) we have a horizontal shift right 5 and an upward shift 3, so the graph is a parabola opening upward with

– Vertex at \((5,3)\) (minimum point),
– Axis of symmetry \(x=5\),
– No real x-intercepts (since \((x-5)^2+3=0\) gives \((x-5)^2=-3\)),
– y-intercept \(f(0)=(0-5)^2+3=28\).

Thus pick the graph whose parabola has vertex \((5,3)\), opens up, and passes through \((0,28)\).

Detailed Explanation

So, the solution is as follows:

  1. Identify the form and transformations.The function is
    \[
    f(x)=(x-5)^2+3,
    \]
    which is the parent parabola \(y=x^2\) shifted right by 5 units and up by 3 units.
  2. Find the vertex.In the form \((x-h)^2+k\), the vertex is \((h,k)\). Here \(h=5\) and \(k=3\), so the vertex is
    \[
    (5,3).
    \]
  3. Axis of symmetry and direction.The axis of symmetry is the vertical line
    \[
    x=5,
    \]
    and because the coefficient of \((x-5)^2\) is \(+1\), the parabola opens upward (same width as \(y=x^2\)).
  4. Intercepts and sample points.Compute the intercepts and a few points:
    • y-intercept: \(f(0)=(0-5)^2+3=25+3=28\), so \((0,28)\).
    • x-intercepts: solve \((x-5)^2+3=0\Rightarrow (x-5)^2=-3\). No real solutions, so there are no x-intercepts.
    • Nearby symmetric points: \(f(6)=(6-5)^2+3=1+3=4\) and \(f(4)=4\), so points \((6,4)\) and \((4,4)\). Also \(f(7)=7\) and \(f(3)=7\).
  5. Conclusion — which graph to choose.Choose the graph that is an upward-opening parabola with vertex at \((5,3)\), axis \(x=5\), no x-intercepts, and passing through points such as \((6,4)\), \((4,4)\), and \((0,28)\). That graph represents \(f(x)=(x-5)^2+3\).
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FAQ

What exactly is f(x) = ()–x?

It looks like a typo or missing expression. Ask for the missing part. If it meant f(x) = -x, the function is y = -x (line through origin with slope -1).

If the function is f(x) = -x, what does its graph look like?

A straight line through (0,0) with slope -1: for each 1 right, go 1 down. Points: (1,-1), (2,-2), (-1,1).

How do I identify slope and intercept from f(x)=mx+b?

m is slope (steepness, rise/run). b is y-intercept (where line crosses y-axis). For y = -x, m = -1 and b = 0.

How do I find x- and y-intercepts?

y-intercept: set x=0 → y=b. X-intercept: set y=0 → solve 0 = mx + b → x = -b/m (if m ≠ 0). For y=-x, both intercepts are 0.

How can I quickly sketch a linear function?

Plot the y-intercept, then use the slope as rise/run to get another point, draw the line through them. Two points determine the line.

How do I check which of several graphs matches the function?

Compare slope and intercept, then test one or two specific x values (e.g., x=1, x=-1) to see if plotted points match.

What is the domain and range of y = -x?

Domain: all real numbers. Range: all real numbers. A non-vertical linear function covers every real y.

Is y = -x symmetric?

It is symmetric about the origin (odd function). Rotating the graph 180° about the origin yields the same line.

How does a graph change if I add/subtract terms (transformations)?

Adding/subtracting constants shifts up/down. Adding/subtracting inside argument shifts left/right. Multiplying by -1 reflects across x-axis. Multiplying by a factor stretches/compresses vertically.

Could the missing parentheses mean absolute value, e.g. f(x)=|x|-x?

Possibly. |x|-x equals 0 for x≥0 and -2x for x<0, producing a piecewise graph: horizontal y=0 for x≥0, line slope -2 for x<0.

How do I recognize non-linear options among graphs?

Look for curves, bends, parabolas, V-shapes, asymptotes, or changes in slope. Linear functions are straight lines.

What is the vertical line test and why use it?

A graph represents a function if every vertical line intersects it at most once. Use it to discard non-function graphs among choices.

Common mistakes when matching function to graph?

Misreading sign of slope, ignoring intercept, testing only one point, or misinterpreting absolute-value/piecewise behavior.

If unsure, what quick check should I do?

Plug in two simple x-values (0 and 1) into the formula and see which graph has those y-values at those x-locations.
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