Repost 8th grade 50 POINTS+BRAINLIEST: How do you graph a quadratic equation in vertex form when the equation is a perfect squar
2 answers:
Answer: Vertex : Maximum (2, 0)
Rules:
- (x + d)² = x² + 2dx + d² and (x - d)² = x² - 2dx + d²
- x² + 2dx = (x + d)² - d² and x² - 2dx = (x - d)² - d²
Solve:
x² - 4x + 4
x² - 2(2x) + 2²
(x - 2)²
Into vertex form: a(x - h)² + k
1(x - 2)² + 0
Identify:
vertex : (h, k) = (2, 0)
Find additional things, to graph the equation:
(i) x-intercept: (2, 0)
(ii) y-intercept: (0, 4)
Graph shown:
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The only point in the solution space is that of selection ...
C (1, 3)
C (1, 3)
Answer:
Since g(x) = f(2x), this means we substitute 2x in place of x for f(x):
g(x) = f(2x) = (2x)² = 4x²
Graphing this, the parabola is not moved from the parent graph f(x) = x², but the graph is compressed horizontally by a factor of 4.Step-by-step explanation:
