Answer
397.5k+ views
Hint:
Here, we will compare the given equation with the general equation of parabola to find the vertex. Then we will find \[x\] and \[y\] intercepts and also an additional point by substituting different values of the variables in the vertex form of the given equation. Using this we will plot those points on a graph and join them together to find the required graph of the given parabola.
Complete step by step solution:
The given quadratic equation in vertex form is $y = - 3{\left( {x - 2} \right)^2} + 5$
Comparing this with $y = a{\left( {x - h} \right)^2} + k$, where $a$ is the axis of symmetry and $\left( {h,k} \right)$ is the vertex.
Now, in order to graph a parabola of this quadratic equation, first of all, we will find the vertex.
Now, since, according to the question, $a = - 3 < 0$ therefore, the vertex will be at the maximum point and the parabola will open downwards.
Thus, comparing the given equation with the general equation, we have,
Vertex $\left( {h,k} \right) = \left( {2,5} \right)$
Now, substituting $x = 0$ in the given quadratic equation $y = - 3{\left( {x - 2} \right)^2} + 5$, we get,
$ y = - 3{\left( {0 - 2} \right)^2} + 5$
$ \Rightarrow y = - 3 \times 4 + 5 = - 12 + 5 = - 7$
Hence, when $x = 0$, $y = - 7$.
Therefore, the $y$-intercept is $\left( {0, - 7} \right)$
Similarly, substituting $y = 0$ in the given quadratic equation $y = - 3{\left( {x - 2} \right)^2} + 5$, we get,
$ 0 = - 3{\left( {x - 2} \right)^2} + 5$
$ \Rightarrow 3{\left( {x - 2} \right)^2} = 5$
Dividing both sides by 3, we get
$ \Rightarrow {\left( {x - 2} \right)^2} = \dfrac{5}{3}$
Taking square root on both sides, we gt
$ \Rightarrow x - 2 = \pm \sqrt {\dfrac{5}{3}} $
Adding 2 on both sides, we get,
$ \Rightarrow x = 2 \pm \dfrac{{\sqrt 5 }}{{\sqrt 3 }}$
Hence, when $y = 0$, $x = 2 + \dfrac{{\sqrt 5 }}{{\sqrt 3 }},2 - \dfrac{{\sqrt 5 }}{{\sqrt 3 }}$
Therefore, the $x$-intercepts are $\left( {2 + \dfrac{{\sqrt 5 }}{{\sqrt 3 }},0} \right)$ and $\left( {2 - \dfrac{{\sqrt 5 }}{{\sqrt 3 }},0} \right)$
If we do the approximate then, the $x$-intercepts are $\left( { \approx 3.291,0} \right)$ and $\left( {0.709,0} \right)$
Now, we will find an additional point also.
Thus, let $x = 4$
Hence, we get,
$y = - 3{\left( {4 - 2} \right)^2} + 5$
$ \Rightarrow y = - 3 \times 4 + 5 = - 12 + 5 = - 7$
Hence, additional point is $\left( {4, - 7} \right)$
Now, we will draw the graph of parabola, such that,
Its vertex is $\left( {2,5} \right)$
The $x$-intercepts are $\left( { \approx 3.291,0} \right)$ and $\left( {0.709,0} \right)$
The $y$-intercept is $\left( {0, - 7} \right)$
An additional point is $\left( {4, - 7} \right)$
Hence, the required graph is:
Hence, this is the required answer.
Note:
A parabola is a curve having a focus and a directrix, such that each point on parabola is at equal distance from them. The axis of symmetry of a parabola is a line about which the parabola is symmetrical. When the parabola is vertical, the line of symmetry is vertical. When a quadratic function is graphed in the coordinate plane, the resulting parabola and corresponding axis of symmetry are vertical.
Here, we will compare the given equation with the general equation of parabola to find the vertex. Then we will find \[x\] and \[y\] intercepts and also an additional point by substituting different values of the variables in the vertex form of the given equation. Using this we will plot those points on a graph and join them together to find the required graph of the given parabola.
Complete step by step solution:
The given quadratic equation in vertex form is $y = - 3{\left( {x - 2} \right)^2} + 5$
Comparing this with $y = a{\left( {x - h} \right)^2} + k$, where $a$ is the axis of symmetry and $\left( {h,k} \right)$ is the vertex.
Now, in order to graph a parabola of this quadratic equation, first of all, we will find the vertex.
Now, since, according to the question, $a = - 3 < 0$ therefore, the vertex will be at the maximum point and the parabola will open downwards.
Thus, comparing the given equation with the general equation, we have,
Vertex $\left( {h,k} \right) = \left( {2,5} \right)$
Now, substituting $x = 0$ in the given quadratic equation $y = - 3{\left( {x - 2} \right)^2} + 5$, we get,
$ y = - 3{\left( {0 - 2} \right)^2} + 5$
$ \Rightarrow y = - 3 \times 4 + 5 = - 12 + 5 = - 7$
Hence, when $x = 0$, $y = - 7$.
Therefore, the $y$-intercept is $\left( {0, - 7} \right)$
Similarly, substituting $y = 0$ in the given quadratic equation $y = - 3{\left( {x - 2} \right)^2} + 5$, we get,
$ 0 = - 3{\left( {x - 2} \right)^2} + 5$
$ \Rightarrow 3{\left( {x - 2} \right)^2} = 5$
Dividing both sides by 3, we get
$ \Rightarrow {\left( {x - 2} \right)^2} = \dfrac{5}{3}$
Taking square root on both sides, we gt
$ \Rightarrow x - 2 = \pm \sqrt {\dfrac{5}{3}} $
Adding 2 on both sides, we get,
$ \Rightarrow x = 2 \pm \dfrac{{\sqrt 5 }}{{\sqrt 3 }}$
Hence, when $y = 0$, $x = 2 + \dfrac{{\sqrt 5 }}{{\sqrt 3 }},2 - \dfrac{{\sqrt 5 }}{{\sqrt 3 }}$
Therefore, the $x$-intercepts are $\left( {2 + \dfrac{{\sqrt 5 }}{{\sqrt 3 }},0} \right)$ and $\left( {2 - \dfrac{{\sqrt 5 }}{{\sqrt 3 }},0} \right)$
If we do the approximate then, the $x$-intercepts are $\left( { \approx 3.291,0} \right)$ and $\left( {0.709,0} \right)$
Now, we will find an additional point also.
Thus, let $x = 4$
Hence, we get,
$y = - 3{\left( {4 - 2} \right)^2} + 5$
$ \Rightarrow y = - 3 \times 4 + 5 = - 12 + 5 = - 7$
Hence, additional point is $\left( {4, - 7} \right)$
Now, we will draw the graph of parabola, such that,
Its vertex is $\left( {2,5} \right)$
The $x$-intercepts are $\left( { \approx 3.291,0} \right)$ and $\left( {0.709,0} \right)$
The $y$-intercept is $\left( {0, - 7} \right)$
An additional point is $\left( {4, - 7} \right)$
Hence, the required graph is:
![seo images](https://www.vedantu.com/question-sets/19ee46a6-38d7-4669-89be-a7895b88f66b7522543296974910652.png)
Hence, this is the required answer.
Note:
A parabola is a curve having a focus and a directrix, such that each point on parabola is at equal distance from them. The axis of symmetry of a parabola is a line about which the parabola is symmetrical. When the parabola is vertical, the line of symmetry is vertical. When a quadratic function is graphed in the coordinate plane, the resulting parabola and corresponding axis of symmetry are vertical.
Recently Updated Pages
Mark and label the given geoinformation on the outline class 11 social science CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
When people say No pun intended what does that mea class 8 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Name the states which share their boundary with Indias class 9 social science CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Give an account of the Northern Plains of India class 9 social science CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Change the following sentences into negative and interrogative class 10 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Advantages and disadvantages of science
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Trending doubts
Bimbisara was the founder of dynasty A Nanda B Haryanka class 6 social science CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Which are the Top 10 Largest Countries of the World?
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
10 examples of evaporation in daily life with explanations
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Give 10 examples for herbs , shrubs , climbers , creepers
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
How do you graph the function fx 4x class 9 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Difference Between Plant Cell and Animal Cell
![arrow-right](/cdn/images/seo-templates/arrow-right.png)