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Chapter 43: Chapter 43

Problem 752

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Find the equation of a parabola that has vertex at \((-1,2)\), axis of symmetry parallel to the \(\mathrm{x}\) -axis, and goes through the point \(\mathrm{P}_{1}(-3,-4)\).

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The equation of the parabola is \(y = -\frac{3}{2}(x+1)^2 +2\).
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TABLE OF CONTENTS :
TABLE OF CONTENTS

Step 1: Write the vertex form of the parabola equation using the given vertex coordinates.

We are given the vertex coordinates (-1, 2). So, the vertex form of the parabola equation will be: \(y = a(x-(-1))^2+2\) Simplify the equation: \(y = a(x+1)^2 +2\)

Step 2: Use the point P1 to find the value of 'a'.

We know the parabola goes through the point P1(-3, -4). We can substitute these coordinates into the equation we found in Step 1 and solve for "a": \(-4 = a(-3+1)^2 +2\)

Step 3: Solve for 'a'.

Simplify the equation and solve for "a": \(-4 = a(-2)^2 +2\) \(-4 = 4a +2\) Subtract 2 from both sides: \(-6 = 4a\) Now, divide by 4: \(a = -\frac{3}{2}\)

Step 4: Write the equation of the parabola.

Now that we have the value of "a", we can write the final equation of the parabola: \(y = -\frac{3}{2}(x+1)^2 +2\) This is the equation of the parabola with vertex at (-1, 2), axis of symmetry parallel to the x-axis, and going through the point P1(-3, -4).

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Most popular questions from this chapter

Chapter 43

Show informally that \(\mathrm{y}=\pm(\mathrm{b} / \mathrm{a}) \mathrm{x}\) are the equations of the asymptotes of the hyperbola whose equation is $$ \left(x^{2} / a^{2}\right)-\left(y^{2} / b^{2}\right)=1 $$
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