Finding the Vertex and end Behavior
- Find the vertex (by using the vertex formula), end behavior, axis of symmetry, x-intercepts(s) and y-intercept(s), given the quadratic equation. Show all your work. You do not need to graph the function.
H(x) = -x² + 10x – 19
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2. Find the degree, end behavior, x- and y intercepts, zeroes of multiplicity, and a few midinterval points of the function. You do not need to graph the function
g(x) = (3x) (x+1) ³
Please do on seperate paper and please show work
The quadratic function H(x) = -x² + 10x – 19 is in standard form, f(x) = ax² + bx + c, where a = -1, b = 10, and c = -19. The vertex formula gives the x-coordinate of the vertex as x = -b/2a. Substituting the values of a and b, we get:
x = -b/2a = -10/(2*(-1)) = 5
To find the y-coordinate of the vertex, we substitute x = 5 in the function H(x):
H(5) = -5² + 10(5) – 19 = 6
Therefore, the vertex of the parabola is (5, 6).
The leading coefficient of the quadratic function is negative, which means that the parabola opens downwards. Therefore, the end behavior of the function is that as x approaches positive or negative infinity, the value of H(x) approaches negative infinity.
The axis of symmetry is a vertical line that passes through the vertex of the parabola. The equation of the axis of symmetry is x = 5.
To find the x-intercepts, we set H(x) = 0 and solve for x:
-x² + 10x – 19 = 0
Using the quadratic formula, we get:
x = [ -10 ± sqrt(10² – 4(-1)(-19)) ] / (2(-1)) x = [ -10 ± sqrt(156) ] / (-2) x = 5 ± sqrt(39)
Therefore, the x-intercepts of the parabola are (5 + sqrt(39), 0) and (5 – sqrt(39), 0).
To find the y-intercept, we set x = 0 in the function H(x):
H(0) = -0² + 10(0) – 19 = -19
Therefore, the y-intercept of the parabola is (0, -19).
