Explore a collection of parabola problems. Detailed solutions with explanations are also provided to help deepen your understanding.
Find the x and y intercepts, the vertex and the axis of symmetry of the parabola with equation \( y = - x^2 + 2 x + 3 \)?
The x-intercepts are the intersection of the parabola with the x-axis, which are points on the x-axis and therefore their y-coordinates are equal to 0. Therefore, we must solve the equation: \[ 0 = - x^2 + 2x + 3 \] Factor the right side of the equation: \[ -(x - 3)(x + 1) = 0\] Solve for x to find: \[ x = 3 \quad \text{and} \quad x = -1\] , The y-intercepts are the intersection of the parabola with the y-axis, which is a point on the y-axis and, therefore, its x-coordinates equal 0 the y-intercept is :\( y = - (0)^2 + 2 (0) + 3 = 3 \), The vertex is found by writing the equation of the parabola in vertex form \(y = a(x - h)^2 + k \) by completing the square and identifying the coordinates of the vertex \( h \) and \( k \) . Complete the square: \[ y = - x^2 + 2 x + 3 = -( x^2 - 2 x - 3) = -( (x - 1)^2 - 1 - 3) = -(x - 1 )^2 + 4 \] Vertex at point \[ (1 , 4) \] You can check all the above points found using the graph of \[ y = - x^2 + 2 x + 3 \] shown below.
What are the points of intersection of the line with equation \( 2x + 3y = 7 \) and the parabola with equation \( y = - 2 x^2 + 2 x + 5\)?
The points of intersection are solutions of the simultaneous equations \[ 2x + 3y = 7 \quad \text{and} \quad y = - 2 x^2 + 2 x + 5 \]. Substitute \( - 2 x^2 + 2 x + 5 \) for \(y \) in the equation \( 2x + 3y = 7 \) to obtain \[ 2x + 3(- 2x^2 + 2x + 5) = 7 \] Write the quadratic equation obtained above in standard form \[ -6x^2 + 8x + 8 = 0 \] Divide all the terms in the equation by 2. \[ -3x^2 + 4x + 4 = 0 \] Solution for x \[ x = 2 \quad , \quad x = -2/3 \] Substitute x for the previous solutions in \( 2x + 3y = 7 \) to find y. \[ x = 2 , y = 1 \quad \text{and} \quad x = -\frac{2}{3}, \quad y = \frac{25}{9} \] The intersection points are: \[ (2 , 1) \quad \text{and} \quad (-\frac{2}{3} , \frac{25}{9} ) \]. Check the answer graphically below.
Find the points of intersection of the two parabolas with equation \( y = -(x - 3)^2 + 2\) and \( y = x^2 - 4x + 1\).
The points of intersection of the two parabolas are solutions of the simultaneous equations
\[ y = -(x - 3)^2 + 2 \quad and \quad y = x^2 - 4x + 1 \].
Substitute \( y \) by \( -(x - 3)^2 + 2 \) in the second equation:
\[ -(x - 3)^2 + 2 = x^2 - 4x + 1 \]
Expand, group like terms and write in standard form
\[ -2x^2 + 10x - 8 = 0 \]
Divide all terms by \( -2 \)
\[ x^2 - 5x + 4 = 0 \]
The solutions of the above quadratic equation are:
\[ x = 1 \quad and \quad x = 4 \]
Use one of the equations to find y:
\( x = 1 \) into the equation \(y = -(x - 3)^2 + 2 \) to get \( y = -(1 - 3)^2 + 2 = -2 \)
\( x = 4 \) into the equation \( y = -(x - 3)^2 + 2 \) to get \( y = -(4 - 3)^2 + 2 = 1 \)
Points of intersection are: \[ (1 , -2) \quad \text{and} \quad (4 , 1) \] Check the answer graphically below.
Find the equation of the parabola \( y = 2 x^2 + b x + c\) that passes through the points \( (-1,-5)\) and \( (2,10)\).
The points \((-1,-5)\) and \((2,10) \) are on the graph of the parabola \( y = 2 x^2 + b x + c\) and therefore are solutions to the equation of the parabola. Substituting by the coordinates of the two points, we obtain the equations:
\[ -5 = 2 (-1)^2 + b (-1) + c\]
and
\[10 = 2 (2)^2 + b (2) + c\]
Rewrite the above system with unknowns \( b \) and \( c \) in standard form.
\[ - b + c = - 7\]
\[ 2b + c = 2\]
Solve the above system of equations to obtain: \( c = - 4 \) and \( b = 3\)
Equation of the parabola that passes through the points \( (-1,-5)\) and \( (2,10)\) is given by: \[ y = 2 x^2 + b x + c = 2 x^2 + 3 x - 4\]
Use a graph plotter to check your answer by graphing \( y = 2 x^2 + 3 x - 4 \) and Check that the graph passes through the points \( (-1,-5) \) and \((2,10)\).
What is the equation of the parabola with x intercepts at \( x = 2\) and \( x = -3\), and a y - intercept at \( y = 5\)?
The equation of a parabola with x-intercepts at \( x = 2 \) and \( x = -3 \) can be written as the product of two factors whose zeros are the x-intercepts as follows: \[ y = a(x - 2)(x + 3) \] We now use the y-intercept at (0, 5), which is a point through which the parabola passes, to write: \[ 5 = a(0 - 2)(0 + 3) \] Solve for \(a\) \[ a = - \dfrac{5}{6} \] Equation: \[ y = - \dfrac{5}{6} (x - 2)(x + 3)\] Graph \( y = - \dfrac{5}{6} (x - 2)(x + 3)\) and verify that the graph has x-intercepts at \( x = 2 , x = -3 \) and a y-intercept at \( y = 5\).
Find the equation of the parabola \( y = a x^2 + b x + c \) that passes through the points \( (0,3) \) , \( (1,-4)\) and \( (-1 , 4)\).
The points \( (0,3), (1,-4) \) and \( (-1,4) \) lie on the graph of the parabola \( y = a x^2 + b x + c \) and are therefore solutions to the equation of the parabola. Therefore, we write the system of 3 equations as follows: \[ (0,3) \Rightarrow c = 3 \quad (I) \] \[ (1,-4) \Rightarrow a + b + 3 = -4 \Rightarrow a + b = -7 \quad (II) \] \[ (-1,4) \Rightarrow a - b + 3 = 4 \Rightarrow a - b = 1 \quad (III) \] Equation (I) gives: \[ c = 3 \] Substitute 3 for c in equations (II) and (III) \[ a + b = -7 \] \[ a - b = 1 \] Solve the system for \( a \) and \( b \) to obatain \[a = - 3 \; , \; b = - 4 \] The equation of the parabola is given by: \[ y = a x^2 + b x + c = -3 x^2 - 4x + 3 \] Graph the graphs of \( y = -3 x^2 - 4x + 3 \) and verify that the graph passes through the points \( (0,3), (1,-4) \) and \( (-1 ,4) \).
Find the equation of the parabola, with vertical axis of symmetry, which is tangent to the line \( y = 3 \) at \( x = -2 \) and its graph passes through the point \((0,5) \ ).
The equation of the parabola, with vertical axis of symmetry, has the form \( y = a x^2 + b x + c \) or in vertex form \( y = a(x - h)^2 + k \) where the vertex is at the point \( (h , k)\) .
In this case it is tangent to a horizontal line \( y = 3 \) at \( x = -2 \) which means that its vertex is at the point \( (h , k) = (-2 , 3) \). Therefore, the equation of this parabola can be written as:
\[ y = a(x - h)^2 + k = a(x - (-2))^2 + 3 = a(x + 2)^2 + 3 \]
Its graph passes through the point \( (0 , 5) \) which must satisfy the equation:
\[ 5 = a(0 + 2)^2 + 3 = 4 a+ 3 \]
Solve the above for \( a \)
\[ a = \dfrac{1}{2} \]
Equation of the parabola is given by: \[ y = \dfrac{1}{2}(x + 2)^2 + 3 \]
Sketch the graphs of \( y = \dfrac{1}{2} (x + 2)^2 + 3 \) and verify that the graph is tangent to the horizontal line \( y = 3 \) at \( x = -2 \) and also the graph passes through the point \( (0 , 5) \).
For what value of the slope m is the line, of equation \( y = m x - 3 \) is tangent to the parabola of equation \( y = 3 x^2 - x \)?
A line and a parabola are tangent if they have one point of intersection only , which is the point of tangency. The points of intersection are found by solving the system \( y = m x - 3 \quad and \quad y = 3 x^2 - x \) Substitute \( y \) by \( m x - 3 \) in the second equation: \[ mx - 3 = 3 x^2 - x \] Write as a standard quadratic equation: \[ 3 x^2 - x(1 + m) + 3 = 0 \] The discriminant of the above quadratic equation is given by: \[ \Delta = (1 + m)^2 - 4(3)(3) \] The line is tangent to the parabola if they have one point of intersection only, which means if: \[\Delta = 0 \] Hence the equation: \[(1 + m)^2 - 4(3)(3) = 0 \] solve for \( m \) \[(1 + m)^2 = 36 \] Solutions are: \[ m = 5 \quad and \quad m = -7 \] Use a graph plotter to check your answer by plotting the graphs of the lines: \( y = 5 x - 3 \) ( \( m = 5 \) solution ), \( y = -7 x - 3 \) ( \( m = 7 \) solution) and the parabola \( y = 3 x^2 - x\) and check that the two lines are tangent to the graph of the parabola \( y = 3 x^2 - x\).
For what values of parameter \( b \) does the line of equation \( y = 2 x + b \) intersect the parabola of equation \( y = - x^2 - 2 x + 1\) at two different points?
The points of intersection are found by solving the system \[ y = 2 x + b \quad and \quad y = - x^2 - 2x + 1 \] Substitute \( y \) by \( 2 x + b \) in the second equation \[ 2x + b = - x^2 - 2x + 1 \] Write as a standard quadratic equation: \[ - x^2 - 4x + 1 - b = 0 \] The discriminant of the above equation is given by: \[ \Delta = (-4)^2 - 4(-1)(1 - b) = 20 - 4b \] The graphs of \( y = 2 x + b \) and \( y = - x^2 - 2 x + 1 \) have two points of intersection if \( \Delta \gt 0 \) (case of two real solutions of a quadratic equation), hence the inequality \[ 20 - 4 b \gt 0 \] Solve for \( b \) \[ b \lt 5 \] Use a graph plotter to check your answer by plotting graphs of \( y = - x^2 - 2 x + 1 \) and lines through equations \( y = 2 x + b \) for values of \( b \gt 5 \) , \( b \lt 5 \) and \( b = 5 \) to see how many points of intersection of the parabola and the line there are for each of these values of \( b \).
Find the equation \( y = a x^2 + x \) of the parabola tangent to the line of equation \( y = 3 x + 1\).
The points of intersection are found by solving the system \[ y = a x^2 + x \quad and \quad y = 3 x + 1 \] Substitute \( y \) by \( 3 x + 1 \) in the equation \( y = a x^2 + x \) \[ 3 x + 1 = a x^2 + x \] Write as a standard quadratic equation: \[ a x^2 - 2 x - 1 = 0 \] Discriminant of the quadratic equation: \[ \Delta = (-2)^2 - 4(a)(-1) = 4 + 4 a \] The graphs are tangent if they have a point of intersection (case for a solution of a quadratic equation) if \( \Delta = 0 \). Hence \[ 4 + 4 a = 0 \] Solve for \(a\) \[ a = -1 \] Parabola equation: \[ y = -x^2 + x \] Graph \( y = - x^2 + x \) and \( y = 3 x + 1 \) to verify the answer above.
Shift the graph of the parabola \( y = x^2 \) to the left 3 units, then reflect the resulting graph in the x-axis, and then shift it up 4 units. What is the equation of the new parabola after these transformations?
Start with: \[ y = x^2 \] Shift 3 units to the left: \[ y = (x + 3)^2 \] Reflect about the x-axis: \[ y = -(x + 3)^2 \] Shift up 4 units: \[ y = -(x + 3)^2 + 4 \]
What transformations are needed to transform the graph of the parabola \( y = x^2 \) into the graph of the parabola \( y = - x^2 + 4 x + 6 \)?
Given: \[ y = - x^2 + 4 x + 6 \] Rewrite in vertex form by completing the square: \[ y = - x^2 + 4 x + 6 = - (x - 2)^2 + 10\] Beginning: \[ y = x^2\] Shift 2 units to the right: \[ y = (x - 2)^2\] Reflect about the x-axis: \[ y = -(x - 2)^2 \] Shift up 10 units: \[ y = -(x - 2)^2 + 10 \]
Write the equation of the parabola shown in the graph below.
Any point identified on the given graph can be used to find the equation of the parabola. However, using the \( x \) and \( y \) intercepts and the vertex are better ways to find the equation of the parabola whose graph is shown below.
Two methods are presented to solve the problem:
method 1:
The graph has two x-intercepts: (-5, 0) and (-1, 0)
Use the two x-intercepts at (-5, 0) and (-1, 0) to write the equation of the parabola as a product of two linea factors:
\[ y = a(x + 1)(x + 5)\]
Use the y-intercept at (0, -5) to write
\[ - 5 = a(0 + 1)(0 + 5) = 5 a\]
Solve for \(a \)
\[ a = -1 \]
Write the equation of the parabola:
\[ y = -(x + 1)(x + 5) = - x^2 - 6 x - 5 \]
method 2:
Use the vertex at \( ( h , k) = (-3 , 4) \) to write the equation of the parabola in vertex form as follows: \[ y = a(x - h)^2 + k = a(x + 3)^2 + 4 \] Use the y-intercept (0, -5) to find \(a\). \[ - 5 = a(0 + 3)^2 + 4 \] Solve the above for \(a\): \[ a = -1 \] The equation of the parabola is given by \[ y = -(x + 3)^2 + 4 = - x^2 -6 x - 5 \]