What is the equation of a line with a slope of −4 and a point (−2, 5) on the line?
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Answer:
x -intercepts are x = -3 and x = 1
y-intercept at y = -3
The line of symmetry is x = -1
vertex is (h,k) = (-1,-4)
Step-by-step explanation:
Given a quadratic equations: h(x) = (x + 1)^2 - 4
The vertex for is :
h(x) = a(x - h)² + k
where,'h' is the axis of symmetry and (h,k) is the vertex.
So from the given equation we will rewrite the equation as:
h(x) = (x - (-1))^2 - 4
Hence,h = -1 and k = -4
The line of symmetry is x = -1 and vertex is (h,k) = (-1,-4)
Now, we have to find the x intercepts,
using the equation,
(x + 1)² = 4
Taking the square root on both sides,
√(x + 1)² = √4
x + 1 = ± 2
x = 1 or x = -3
So x -intercepts are x = -3 and x = 1
For y-intercept put x = 0 into the real equation:
h(x) = (0 +1)^2 - 4
y = 1 - 4
y = -3
So y-intercept at y = -3
Here with I have attached the graph.
Thank you.
