Solveeit Logo
Login

Question

Question: How do you find the x-intercepts of the parabola with vertex (4, -1) and y-intercept of (0, 15)?...

How do you find the x-intercepts of the parabola with vertex (4, -1) and y-intercept of (0, 15)?

Explanation

Solution

In the given question we have been asked to find the value of x-intercept. We have given the vertex of the given parabola by using the vertex form of the equation; i.e. y=a(x−h)2+ky=a{{\left( x-h \right)}^{2}}+k where, aa equals to the coefficient of x2{{x}^{2}}, ‘h’ is the x-coordinate of the vertex, ‘k’ is the y-coordinate of the vertex, we will formed the equation of the parabola. Then by using the (0, 15) we will find the value of ‘a’. Later for finding the x-intercept, we will need to put y = 0 and we will get the value of x-intercept.

Complete step by step answer:
The equation of the parabola is,
y=a(x−h)2+ky=a{{\left( x-h \right)}^{2}}+k Where, aa equals to the coefficient of x2{{x}^{2}}, ‘h’ is the x-coordinate of the vertex, ‘k’ is the y-coordinate of the vertex.
We have given the vertex of the parabola i.e. (4, -1) = (h, k)
Equation formed;
y=a(x−4)2−1y=a{{\left( x-4 \right)}^{2}}-1
We have given the y-intercept i.e. the parabola passes through the point (0, 15).
Thus,
The point (0, 15) will satisfy the parabola equation.
Putting x = 0 and y = 15 in the above equation, we get
15=a(0−4)2−115=a{{\left( 0-4 \right)}^{2}}-1
Simplifying the above expression, we get
15=16a−115=16a-1
Adding 1 to both the sides of the equation, we get
16=16a16=16a
Dividing both the sides by 16, we get
a=1a=1
Therefore,
The equation of parabola is;
y=1(x−4)2−1y=1{{\left( x-4 \right)}^{2}}-1
Simplifying the above equation by using the identity i.e. (a−b)2=a2−2ab+b2{{\left( a-b \right)}^{2}}={{a}^{2}}-2ab+{{b}^{2}}
y=x2−8x+16−1y={{x}^{2}}-8x+16-1
⇒y=x2−8x+15\Rightarrow y={{x}^{2}}-8x+15
Now,
Finding the x-intercept,
We need to put the value of y = 0,
We have,
y=x2−8x+15y={{x}^{2}}-8x+15
⇒0=x2−8x+15\Rightarrow 0={{x}^{2}}-8x+15
Rewrite the above equation as,
x2−8x+15=0{{x}^{2}}-8x+15=0
The quadratic formula provides the solution for the quadratic equation:
ax2+bx+c=0a{{x}^{2}}+bx+c=0
In which a, b and c are the coefficient of respectively terms in the quadratic equation, as follows:
Roots of the quadratic equation= −b±b2−4ac2a\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}
Determine the quadratic equation’s coefficients a, b and c:
The coefficient of the given quadratic equation x2−8x+15=0{{x}^{2}}-8x+15=0are,
a=1 ⇒b=−8 ⇒c=15a = 1\\\ \Rightarrow b = -8\\\ \Rightarrow c = 15
Plug these coefficient into the quadratic formula:
−b±b2−4ac2a=−(−8)±(−8)2−(4×1×15)2\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}=\dfrac{-\left( -8 \right)\pm \sqrt{{{\left( -8 \right)}^{2}}-\left( 4\times 1\times 15 \right)}}{2}
Solve exponents and square root, we get
⇒8±64−(4×1×15)2\Rightarrow \dfrac{8\pm \sqrt{64-\left( 4\times 1\times 15 \right)}}{2}
Performing any multiplication and division given in the formula,
⇒8±64−(60)2\Rightarrow \dfrac{8\pm \sqrt{64-\left( 60 \right)}}{2}
⇒8±42=8±22\Rightarrow \dfrac{8\pm \sqrt{4}}{2}=\dfrac{8\pm 2}{2}
We got two values, i.e.
⇒8+22 and 8−22\Rightarrow \dfrac{8+2}{2}\, and\, \dfrac{8-2}{2}
Solving the above, we get
⇒102 and 62\Rightarrow \dfrac{10}{2}\, and\, \dfrac{6}{2}
Converting into simplest form,
⇒5 and 3\Rightarrow 5\, and\, 3
Therefore,
∴x=14,−3\therefore x=\dfrac{1}{4},-3

Therefore,the possible value of x-intercept is 3 and 53\,and\, 5.

Note: While solving these types of questions, students need to know the concept of finding intercept. Solve the equation very carefully and do the calculation part very explicitly to avoid making any errors. They should be well aware about the concept of equation of the parabola and a way an equation formed using the given vertex and a point by which a given line passes through.