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write the quadratic function of the parabola that has a vertex at (4, 3…

Question

write the quadratic function of the parabola that has a vertex at (4, 3) and goes through point (-5, -6) by filling in the value of a. (round your answer to the nearest hundredth, if necessary.)
f(x) = a(x - 4)^2 + 3
answer

Explanation:

Step1: Substitute the point into the function

We know the function is \( f(x) = a(x - 4)^2 + 3 \) and it passes through the point \((-5, -6)\). So we substitute \( x=-5 \) and \( f(x)=-6 \) into the function:
\( -6 = a(-5 - 4)^2 + 3 \)

Step2: Simplify the equation

First, calculate \( -5 - 4=-9 \), then square it: \( (-9)^2 = 81 \). So the equation becomes:
\( -6 = 81a + 3 \)

Step3: Solve for \( a \)

Subtract 3 from both sides:
\( -6 - 3 = 81a \)
\( -9 = 81a \)
Then divide both sides by 81:
\( a=\frac{-9}{81}=\frac{-1}{9}\approx - 0.11 \) (rounded to the nearest hundredth)

Answer:

\( a\approx -0.11 \), so the quadratic function is \( f(x)=-0.11(x - 4)^2 + 3 \) (or with the exact fraction, but the question asks for the value of \( a \) rounded to the nearest hundredth, so the value of \( a \) is approximately \(-0.11\))