Where is the removable discontinuity of f(x) located? x = -5 F(x)=5x+10/x^2+7x+10
2 answers:
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Answer:
a = -0.32
b = 16
c = - 168
Step-by-step explanation:
In the figure attached, the quadratic function is shown. We can model it as follows:
y = a*(x - x1)*(x - x2)
where (x1, 0) and (x2, 0) are its roots
From the picture, roots are (15, 0) and (35, 0), replacing them into the equation:
y = a*(x - 15)*(x - 35)
and the vertex is at (25,32), replacing into the equation:
32 = a*(25 - 15)*(25 - 35)
32 = a*(-100)
a = 32/(-100)
a = -0.32
Applying distributive property:
y = -0.32*(x - 15)*(x - 35)
y = -0.32*(x² - 35x - 15x + 525)
y = -0.32x² + 16x - 168
which corresponds to the general form:
y = ax² + bx + c
