Parabola Questions and Problems with Detailed Solutions

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: $$x=3\text{and}x=-1$$ , 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: $$y=-x^2+2x+3=-(x^2-2x-3)=-((x-1{)}^2-1-3)=-(x-1{)}^2+4$$ Vertex at point $$(1,4)$$ You can check all the above points found using the graph of $$y=-x^2+2x+3$$ shown below.

The points of intersection are solutions of the simultaneous equations $$2x+3y=7\text{and}y=-2x^2+2x+5$$. Substitute $-2x^2+2x+5$ 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 $$x=2,x=-2/3$$ Substitute x for the previous solutions in $2x+3y=7$ to find y. $$x=2,y=1\text{and}x=-\frac{2}{3},y=\frac{25}{9}$$ The intersection points are: $$(2,1)\text{and}(-\frac{2}{3},\frac{25}{9})$$. Check the answer graphically below.

The points of intersection of the two parabolas are solutions of the simultaneous equations $$y=-(x-3{)}^2+2andy=x^2-4x+1$$.
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: $$x=1andx=4$$ 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: $$(1,-2)\text{and}(4,1)$$ Check the answer graphically below.

The points $(-1,-5)$ and $(2,10)$ are on the graph of the parabola $y=2x^2+bx+c$ 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: $$y=2x^2+bx+c=2x^2+3x-4$$ Use a graph plotter to check your answer by graphing $y=2x^2+3x-4$ and Check that the graph passes through the points $(-1,-5)$ and $(2,10)$.

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 $a$ $$a=-{\displaystyle \frac{5}{6}}$$ Equation: $$y=-{\displaystyle \frac{5}{6}}(x-2)(x+3)$$ Graph $y=-{\displaystyle \frac{5}{6}}(x-2)(x+3)$ and verify that the graph has x-intercepts at $x=2,x=-3$ and a y-intercept at $y=5$.

The points $(0,3),(1,-4)$ and $(-1,4)$ lie on the graph of the parabola $y=a{x}^2+bx+c$ and are therefore solutions to the equation of the parabola. Therefore, we write the system of 3 equations as follows: $$(0,3)\Rightarrow c=3(I)$$ $$(1,-4)\Rightarrow a+b+3=-4\Rightarrow a+b=-7(II)$$ $$(-1,4)\Rightarrow a-b+3=4\Rightarrow a-b=1(III)$$ 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: $$y=a{x}^2+bx+c=-3x^2-4x+3$$ Graph the graphs of $y=-3x^2-4x+3$ and verify that the graph passes through the points $(0,3),(1,-4)$ and $(-1,4)$.

The equation of the parabola, with vertical axis of symmetry, has the form $y=a{x}^2+bx+c$ 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: $$5=a(0+2{)}^2+3=4a+3$$ Solve the above for $a$ $$a={\displaystyle \frac{1}{2}}$$ Equation of the parabola is given by: $$y={\displaystyle \frac{1}{2}}(x+2{)}^2+3$$ Sketch the graphs of $y={\displaystyle \frac{1}{2}}(x+2{)}^2+3$ and verify that the graph is tangent to the horizontal line $y=3$ at $x=-2$ and also the graph passes through the point $(0,5)$.

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 $y=mx-3andy=3x^2-x$ Substitute $y$ by $mx-3$ in the second equation: $$mx-3=3x^2-x$$ Write as a standard quadratic equation: $$3x^2-x(1+m)+3=0$$ The discriminant of the above quadratic equation is given by: $$\mathrm{\Delta }=(1+m{)}^2-4(3)(3)$$ The line is tangent to the parabola if they have one point of intersection only, which means if: $$\mathrm{\Delta }=0$$ Hence the equation: $$(1+m{)}^2-4(3)(3)=0$$ solve for $m$ $$(1+m{)}^2=36$$ Solutions are: $$m=5andm=-7$$ Use a graph plotter to check your answer by plotting the graphs of the lines: $y=5x-3$ ( $m=5$ solution ), $y=-7x-3$ ( $m=7$ solution) and the parabola $y=3x^2-x$ and check that the two lines are tangent to the graph of the parabola $y=3x^2-x$.

The points of intersection are found by solving the system $$y=2x+bandy=-x^2-2x+1$$ Substitute $y$ by $2x+b$ 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: $$\mathrm{\Delta }=(-4{)}^2-4(-1)(1-b)=20-4b$$ The graphs of $y=2x+b$ and $y=-x^2-2x+1$ have two points of intersection if $\mathrm{\Delta }>0$ (case of two real solutions of a quadratic equation), hence the inequality $$20-4b>0$$ Solve for $b$ $$b<5$$ Use a graph plotter to check your answer by plotting graphs of $y=-x^2-2x+1$ and lines through equations $y=2x+b$ for values of $b>5$ , $b<5$ and $b=5$ to see how many points of intersection of the parabola and the line there are for each of these values of $b$.

The points of intersection are found by solving the system $$y=a{x}^2+xandy=3x+1$$ Substitute $y$ by $3x+1$ in the equation $y=a{x}^2+x$ $$3x+1=a{x}^2+x$$ Write as a standard quadratic equation: $$a{x}^2-2x-1=0$$ Discriminant of the quadratic equation: $$\mathrm{\Delta }=(-2{)}^2-4(a)(-1)=4+4a$$ The graphs are tangent if they have a point of intersection (case for a solution of a quadratic equation) if $\mathrm{\Delta }=0$. Hence $$4+4a=0$$ Solve for $a$ $$a=-1$$ Parabola equation: $$y=-x^2+x$$ Graph $y=-x^2+x$ and $y=3x+1$ to verify the answer above.

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: $$y=-(x+3{)}^2+4$$

Given: $$y=-x^2+4x+6$$ Rewrite in vertex form by completing the square: $$y=-x^2+4x+6=-(x-2{)}^2+10$$ 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: $$y=-(x-2{)}^2+10$$

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 $$-5=a(0+1)(0+5)=5a$$ Solve for $a$ $$a=-1$$ Write the equation of the parabola: $$y=-(x+1)(x+5)=-x^2-6x-5$$

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 $$y=-(x+3{)}^2+4=-x^2-6x-5$$